91. Calculate the [H+] for a 0.10 M acetic acid solution. Then calculate % ionization

1.8 x 10−5 =

x = [ H+ ] = 1.3 x 10-3 M

% ionization =

x 100 = 1.3%

pH = - log [1.3 x 10-3 M] = 2.89

92. Calculate the [H+] for a 0.010 M acetic acid solution. Then calculate % ionization

1.8 x 10−5 =

x = [ H+ ] = 4.2 x 10-4 M

% ionization =

x 100 = 4.2%

pH = - log [4.2 x 10-4 M] = 3.38

93. Calculate the [H+] for a 0.0010 M acetic acid solution. Then calculate % ionization

1.8 x 10−5 =

x = [ H+ ] = 1.3 x 10-4 M

% ionization =

x 100 = 13%

pH = - log [1.3 x 10-4 M] = 3.87

94. Calculate the [H+] for a 0.00010 M acetic acid solution. Then calculate % ionization

1.8 x 10−5 =

x = [ H+ ] = 4.2 x 10-5 M

% ionization =

x 100 = 34%

pH = - log [4.2 x 10-5 M] = 4.47 95. The percent ionization of a particular weak acid increases as the acid

concentration decreases. Note that this does not mean that the [H+] increases or the pH decreases for more dilute solutions of the same acid, because the amount of H+ ions present will be less, but the amount of H+ present will be a bigger percentage of the original, more dilute acid. For HA ⇌H+ + A− when diluting the

solution, there will be less concentration of HA in the water and less concentration of H+ and A− in the water, thus the equilibrium will shift to products because one acid molecule can produce two ions, increasing overall concentration more.

This is analogous to the rational we used to explain why gas equilibria shift to the side with more gas molecules when we increase the volume of the container (decrease pressure). Decreasing concentration is analogous to increasing container size, and the shift is towards more gas molecules, just as the shift here is toward the side with more particles, the ions.

96. Calculate the pH of the solution that results from adding 5 drops (~0.25 ml) of 1.0 M HCl to 50 ml of water.

HCl is a SA, and is being diluted with water, thus this is a simple dilution problem, (don’t forget to use total volume) thus we can use the dilution formula. Since HCl is a strong acid, the [HCl] = [H+]

MdVd = MMVM (1M)(0.25ml ) = (x)(50.25ml ) x = [HCl ] = [H+ ] = 0.00498M

−log[0.00498M] = 2.30

97. Calculate the pH of the solution that results from adding 5 drops (~0.25 ml) of 1.0 M NaOH to 50 ml of water.

NaOH is a SB, and is being diluted with water, thus this is a simple dilution problem, (don’t forget to use total volume) thus we can use the dilution formula. Since NaOH is a strong base, the [NaOH] = [OH−]

MdVd = MMVM (1M)(0.25ml ) = (x)(50.25ml ) x =[NaOH]=[OH−]= 0.00498M

pOH = −log[0.00498M] = 2.30 14.00 - pH = 11.70

98. Write a reaction to represent the dissociation of aqueous acetic acid, HC2H3O2. Write the equilibrium expression.

HC2H3O2(aq) + H2O(L) ⇌ H3O+(aq) + C2H3O2

−(aq) OR

HC2H3O2(aq) ⇌ H+(aq) + C2H3O2

−(aq)

Ka = [H+] [C2H3O2−]

[HC2H3O2]

a. Calculate the pH a 125 ml of vinegar, which is 0.80 M acetic acid, Ka = 1.8 × 10−5.

This is just a WA in water, thus this is a simple “x2” problem for the weak acid vinegar. In the equilibrium expression [H+] = [C2H3O2

−] = [x]

1.8 ×10−5 = [x][x] [x] = [H+] = 0.0038M

[0.8] pH = -log [0.0038M] = 2.42

b. Calculate the pH of 50.0 ml of 0.80 M acetic acid to which 1.64 g of sodium acetate (MM 82 g/mol) has been added. (Assume no volume change occurs when the solid is added.)

This is a simple buffer problem. Nothing added to the buffer (therefore no +/−

incoming), just the WA and WB in the buffer equation: [H+ ] = Ka

We need our WA and WB in either molarity or moles (millimoles).... I recommend millimoles.

Acetic acid is a WA; 0.80 M × 50 ml = 40 mmol of acetic acid

1.64 g of sodium acetate is 0.020 mol, which is 20 mmol of acetate ion, the WB.

substitute and solve [H+ ] = 1.8 ×10−5

= 3.6 ×10−5M

pH = −log[3.6 ×10−5M] = 4.44

c. Calculate the pH of 50.0 ml of 0.80 M acetic acid to which 3.28 g of sodium acetate has been added. (Assume no volume change.)

This is same problem as part. Nothing added to the buffer (therefore no +/−

incoming), just the WA and WB in the buffer equation: [H+ ] = Ka

We need our WA and WB in either molarity or moles (millimoles).... I recommend millimoles.

Acetic acid is a WA; 0.80 M × 50 ml = 40 mmol of acetic acid

3.28 g of sodium acetate is 0.040 mol, which is 40 mmol of acetate ion, the WB. This is what I would call a “half-way” buffer as there are equal quantities of the

WA and is cjWB. Substitute and solve [H+ ] = 1.8 ×10−5

= 1.8 ×10−5M

pH = −log[1.8 ×10−5M] = 4.74

d. Calculate the pH of the resulting solution when 1.0 ml of 1.0 M HCl is added to the solution described in part c.

This is a buffer problem in which there is incoming strong acid.

[H+]= Ka

Again, I recommend you work this problem in moles (milli moles). Incoming strong

base; 1M × 1.5ml = 1mmol. Substitute and solve [H+] = 1.8×10−5

=1.9x10-5

pH = −log[1.9 ×10−5M] = 4.72

I 0.10 ~0 ~0

C -x (too small) +x +x

E 0.10 x x

I 0.010 ~0 ~0

C -x (too small) +x +x

E 0.010 x x

I 0.0010 ~0 ~0

C -x (too small) +x +x

E 0.0010 x x

I 0.00010 ~0 ~0

C -x (too small) +x +x

E 0.00010 x x

e. Calculate the pH of the resulting solution when 2.0 ml of 1.0 M NaOH is added to the solution described in part c.

This is a buffer problem in which there is incoming strong base.

[H+]= Ka

Take a hint, work this problem in moles (milli moles). Incoming strong base; 1M ×

2ml = 2mmol. Substitute and solve [H+] = 1.8×10−5

=1.6x10-5

pH = −log[1.6 ×10−5M] = 4.79

99. Write a reaction to represent the ionization of hypochlorous acid. Write the equilibrium expression.

HClO(aq) + H2O(L) ⇌ H3O+(aq) + ClO−

(aq) OR HClO(aq) ⇌ H+(aq) + ClO−

(aq)

Ka = [H+] [ClO−] [HClO]

a. Calculate the pH of 0.55 M hypochlorous acid. (Ka = 3.0 × 10−8)

This is just a WA in water, thus this is a simple “x2” problem for the weak acid

vinegar. Ka =

In the equilibrium expression [H+] = [ClO−] = [x] 3.0 ×10−8 =

[x] = [H+] = 0.00013M pH = - log[0.00013M] = 3.89

b. Calculate the pH of 100.0 ml of 0.55 M hypochlorous acid to which 3.49 g of potassium hypochlorite (MM 90.55 g/mol) has been added. (Assume no volume change occurs when the solid is added.)

This is a simple buffer problem. Nothing added to the buffer (therefore no +/−

incoming), just the WA and WB in the buffer equation: [H+] = Ka

We need our WA and WB in either molarity or moles (millimoles).... I always recommend millimoles. Hypochlorous acid is a WA; 0.55 M × 100 ml = 55 mmol of hypochlorous acid

3.49 g of sodium acetate is 0.0385 mol, which is 38.5 mmol of hypochlorite ion,

the WB. substitute and solve [H+ ] = 3.0 ×10−8

= 4.2 ×10−8 M

pH = - log [4.2 ×10−8 M] = 7.37

c. Calculate the mass of potassium hypochlorite that must be added to 100.0 ml of 0.55 M hypochlorous acid to produce a solution with a pH of 7.00 (Assume no volume change occurs when the solid is added.)

This is a simple buffer problem, the same as part (b) except this time we know the desired pH which we can put in for [H+], and instead we solve for the millimol of

WB, the hypochlorite ion. [H+] = Ka

Hypochlorous acid is a WA; 0.55 M × 100 ml = 55 mmol of hypochlorous acid. Since pH = 7, we know that [H+] = 1 × 10−7

Substitute and solve [1×10−7 ] = 3.0×10−8

x = 16.5mmol

Thus 16.5 mmol of hypochlorite ion is 0.0165 mol of KClO calculate:

0.0165 mol × 90.55 g/mol = 1.5 g

100. Write a reaction to represent the hydrolysis of methylamine. (CH3NH2, Kb = 4.4 × 10−4) Write the equilibrium expression. Then write the reaction of the hydrolysis that occurs when methylamide chloride (CH3NH3Cl). Write the equilibrium expression and calculate the Ka for this reaction.

CH3NH2(aq) + H2O(L) ⇌ CH3NH3+

(aq) + OH−(aq) Kb = [OH− ][CH3NH3

+]

[CH3NH2]

Next, recognize that this is a chloride the salt which dissolves: CH3NH3Cl(aq) → CH3NH3

+(aq) + Cl−(aq) (note the single arrow)

The methylamide is the conjugate weak acid of the WB, methylamine. This is the equilibrium reaction of methyl amide ion in water:

CH3NH3+

(aq) ⇌ CH3NH2(aq) + H+(aq) Ka = [H+][CH3NH2]

[CH3NH3+]

Ka × Kb = Kw Ka =

=

= 2.3 ×10−11

a. Calculate the pH of 75.0 ml of 0.035 M aqueous methylamine.

This is just a WB in water. Thus this is a simple “x2” problem for the weak base methylamine. In the equilibrium expression [OH−] = [CH3NH3

+] = [x]

4.4 ×10−4 = [ x ][ x ] [x] = [OH−]= 0.0039 M pOH = -log [0.0039M] = 2.41

[0.035] pH = 14.00 – pOH = 11.59

b. Calculate the pH of 0.25 g of methylamide chloride (MM 67.52 g/mol) dissolved to make 75.0 ml of solution.

Reread the initial prompt above to help understand that is a simple “x2” problem for a cjWA, the methylamide ion. As we already calculated the Ka of this conjugate weak acid above…we need a molarity for the methylamide ion, the WA:

0.25 g 1 mol = 0.0037 mol 0.0037 mol = 0.049 M 67.52 g 0.075 L

In this “x2” problem, [H+] = [CH3NH2] = [x]

2.3 x 10-11 =

x = [H+] = 1.06 x 10-6M

pH = −log[11.06 x 10-6M] = 5.97

c. Calculate the pH of 75.0 ml of 0.035 M aqueous methylamine to which 0.25 g of methylamide chloride has been added. (Assume no volume change occurs when the solid is added.)

This is a simple buffer problem. Nothing added to the buffer (therefore no +/− incoming), just the WA and WB in the buffer equation:

[H+] = Ka

OR [OH−] =Kb

We need our WA and WB in either molarity or moles (millimoles).... Can you take my recommendation of millimoles yet?

From part (c), the 0.25 g of methylamide chloride is 3.7 mmol of methylamide ion, the WA.

The WB is the methylamine 0.035 M × 75 ml = 2.625 mmol of methylamine, the WB.

substitute and solve [H+] = 2.3×10−11

= 3.24 ×10−11M

OR [OH−] = 4.4 ×10−4

= 3.09 ×10−4M

In either case pH = 10.49

d. Calculate the pH of methylamine / methylamide chloride buffer that contains equal number of moles of both the weak base and its conjugate weak acid.

This is a simple buffer problem. Nothing added to the buffer, just the WA and WB in the buffer equation.

In this buffer WA = WB, and will cancel out of the equation, thus

[H+] = Ka [H+] = 2.3×10−11 OR [OH−] = Kb [OH−] = 4.4 ×10−4

In either pH = 10.64

e. Calculate the pH of the solution that results when 1.5 ml of 1.0 M HCl is added to the solution described in part c.

This is a buffer problem in which there is incoming strong acid.

[H+] = Ka

OR [OH−] = Kb

Assuming you got the millimoles hint. Incoming strong acid; 1 M × 1.5 ml = 1.5 mmol

Substitute and solve

[H+] = 2.3×10−11

OR [OH−] = 4.4×10−4

pH = 9.97

101. Write a reaction to represent the dissociation of aqueous ascorbic acid, HC6H7O6. Write the equilibrium expression.

Ascorbic acid is a WA. HC6H7O6 (aq) + H2O(L) ⇌ H3O+(aq) + C6H7O6

− (aq) OR

HC6H7O6(aq) ⇌ H+(aq) + C6H7O6

−(aq)

Ka = [H+] [C6H7O6−]

[HC6H7O6]

a. Calculate the pH a 225 ml of an aqueous solution of 0.50 M ascorbic acid, Ka = 8.0 x 10−5.

This is just a WA in water, thus this is a simple “x2” problem for the weak acid ascorbic acid. In the equilibrium expression

[H+] = [C6H7O6−] = [x] 8.0 ×10−5 =

[x] = [H+] = 0.0063M

pH = 2.20

b. Calculate the pH of 100.0 ml of 0.50 M ascorbic acid to which 1.1 g of potassium ascorbate, KC6H7O6 (MM 214.2 g/mol) has been added. (Assume no volume change occurs when the solid is added.)

This is a simple buffer problem. Nothing added to the buffer (therefore no +/−

incoming), just the WA and WB in the buffer equation: [H+ ] = Ka

We need our WA and WB in either molarity or moles (millimoles).... so of course millimoles.

Ascorbic acid is a WA; 0.50 M × 100 ml = 50 mmol of ascorbic acid

1.1 g of potassium ascorbate is 0.0051 mol, which is 5.1 mmol of ascorbate ion, the WB.

substitute and solve [H+ ] = 8.0 ×10−5

= 7.8 ×10−4M pH = 3.12

c. Calculate the mass of potassium ascorbate that must be added to 0.50 M ascorbic acid when making a solution with a volume of 100.0 ml to generate a pH of 4.10

This is a simple buffer problem, the same as part (b). Except this time we know the desired pH which we can put in for [H+], and instead we solbe for the millimol

of WB, the ascorbat ion. [H+ ] = Ka

Ascorbic acid is a WA; 0.50 M × 100 ml = 50 mmol of ascorbic acid . Since pH=4.10, we know that [H+] = 7.9 x 10-5

Substitute and solve [7.9 x 10-5] = 8×10−5

Thus 50.4 mmol of ascorbate ion is 0.0504 mol of KC6H7O6 calculate: 0.0504 mol × 214.2 g/mol = 11 g

d. Calculate the pH of the resulting solution when 1.0 ml of 1.0 M HCl is added to the solution described in part c.

This is a buffer problem in which there is incoming strong acid.

[H+] = Ka

Take the blasted millimoles hint. Incoming strong acid; 1 M × 1 ml = 1 mmol

Substitute and solve

[H+] = 8×10−5

[H+] = 8.3×10−5

pH = 4.08

e. Calculate the pH of the resulting solution when 1.0 ml of 1.0 M NaOH is added to the solution described in part c.

This is a buffer problem in which there is incoming strong base.

[H+] = Ka

Oh millimoles how useful you are... Incoming strong base; 1 M × 1 ml = 1 mmol

Substitute and solve

[H+] = 8×10−5

[H+] = 7.6×10−5

pH = 4.12

102. Is nitric acid a strong or weak acid? Write a reaction that represents its ionization in water.

Nitric is one of the seven strong acids. HNO3(aq) → H+

(aq) + NO3−

(aq) OR HNO3(aq) + H2O(L) → H3O+

(aq) + NO3−

(aq)

103. Is sodium hydroxide a strong or weak base? Write a reaction that represents its ionization in water.

NaOH is a strong base. NaOH(aq) → Na+ + OH−(aq)

104. Determine the initial pH of the 50.0 ml of 0.040 M nitric acid solution.

Since we have a SA which dissociates completely, the [H+] is = the [HNO3], thus pH = −log[H+]

pH = −log(0.040 M) = 1.40

(remember for pH 1.40 has only 2 sig figs, the 1.40, the 1 in front is magnitude, not actually part of the 2 sig figs in 0.040 M.)

105. Determine the initial pH of the 0.025 M sodium hydroxide solution.

Since we have a SB which dissociates completely, the [OH−] is = the [NaOH], thus pOH = −log[OH−]

pOH = −log(0.025M) = 1.60, 14.00 – pOH = pH = 12.40

106. What volume of sodium hydroxide would you need to add to the 50.0 ml of 0.040 M nitric acid if you wanted to completely neutralize this acid?

This is called the equivalence point. Since we have a “monobasic” SB reacting with a monoprotic SA the acid and base react in a 1:1 ratio.

We need to calculate the moment when moles H+ = moles OH−.

We can use the MaVa = MbVb. You can do your calculations in ml, which would mean you would be calculating millimoles instead of moles, but since the volume is in this equation twice, it will simply report the volume in ml (You can use liters

for your volume of acid, it will simply mean your volume of base would come out in liters, 0.080 L).

Thus (0.04)(50) = (0.025)(Vb), and Vb = 80 ml of 0.025 M NaOH

107. What would be the pH of this resulting solution from the previous question?

Always at the equivalence point of a SA with a SB, the pH will be equal to 7. This is because the only species in solution are the pathetic conjugates of the SA and SB.

108. Calculate the pH of the resulting solution when 5.0 ml of the 0.025 M sodium hydroxide is added to 50.0 ml of 0.040 M nitric acid.

Tackling this question is really just a limiting reactant problem. Since you know that it takes 80 ml of base to neutralize, any volume of base up until 80 ml will mean that there is excess acid, and the pH will be a function of the [H+].

Calculate mmol of acid using MV, (0.04M)(50ml) = 2 mmol and mmol base is (0.025M)(5ml) = 0.125 mmol OH−

A simple subtraction leaves 1.875 mmol H+ 1.875mmol H+ = 0.034 M H+ 55ml TotalVol

pH = −log(0.034) = 1.47

109. Calculate the pH of the resulting solution when 40.0 ml of the 0.025 M sodium hydroxide is added to 50.0 ml of 0.040 M nitric acid.

This problem is the same process as question #198. Calculate the incoming mmol base; (0.025M)(40ml) = 1 mmol OH−

A simple subtraction from the previously calculated 2 mmol acid leaves 1 mmol H+. Use this 1 mmol of H+ to calculate the concentration of that excess acid that has not yet been neutralized

1.0mmol H+ = 0.011M H+ pH =−log(0.011) = 1.95

90ml TotalVol

110. Calculate the pH of the resulting solution when 75.0 ml of the 0.025 M sodium hydroxide is added to 50.0 ml of 0.040 M nitric acid.

Same process as problem #108 and #109. Calculate the incoming mmol base is (0.025M)(75ml) = 1.875 mmol OH−

Subtraction from the previously calculated 2 mmol acid leaves 0.125 mmol H+ Use this 0.125 mmol of H+ to calculate the concentration of that excess acid that has not yet been neutralized

0.125mmol H+ = 0.00100 M H+ pH =−log(0.0010) = 3.000

125ml TotalVol

111. Calculate the pH of the resulting solution when 90.0 ml of the 0.025 M sodium hydroxide is added to 50.0 ml of 0.040 M nitric acid.

Now that we have gone beyond the equivalence point, there is 10 ml excess base, since we calculated in #106 that 80 ml of base is required for the neutralization, thus the pH will be completely dependent on the moles of excess base in the total volume of the solution.

mmol excess base is (0.025M)(10ml) = 0.25 mmol OH−. Use this mmol of OH− to calculate the concentration of that excess base thatis now in the solution.

0.25mmol OH− = 0.0018M OH− pOH =−log(0.0018) = 2.74 140ml TotalVol 14.00 – pOH = 11.26

112. Make a sketch of the titration curve using the volumes and pH values calculated for the titration of 30 ml of 0.00 M sodium hydroxide with 0.050 M hydrochloric acid right.

As you construct this graph, you will get a much more accurate sense of the titration curve if you make two more calculations and place two more points on the graph.

113. Calculate the pH of the resulting solution when 23.0 ml of the 0.050 M hydrochloric acid is added to 30.0 ml of 0.040 M potassium hydroxide.

114. Calculate the pH of the resulting solution when 25.0 ml of the 0.050 M hydrochloric acid is added to 30.0 ml of 0.040 M potassium hydroxide.

115. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HC3H5O3.

This is an “x2” problem

Ka = [H+] [A−] = 1.4 x 10-4 = x2 x = [H+] = 5.3 × 10−3 M [HA] [0.20] pH = 2.28

116. What volume of 0.40 M KOH would be needed to neutralize this 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3? (This should be the first step you would take to solve any titration problem, even if you weren’t asked to do it.)

It is ALWAYS important to calculate the volume required to reach the equivalence point; moles of acid = moles of base, MaVa = MbVb = 0.20 M × 50 ml = 0.40 M × Vb

Thus the volume of SB needed to reach the equivalence point is

25.0 ml of 0.40 KOH

117. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with 10.0 ml of 0.40 M KOH?

At this point you are only partway through the titration, but not yet to the halfway point, so use the “buffer equation,” which is just an algebraic rewrite of the equilibrium expression.

Calculate the millimoles of lactic acid (0.2M × 50ml) and millimoles of incoming strong base (0.4M × 10ml)

Subtract the incoming strong base from the weak acid, to find out the remaining un-neutralized weak acid (6 mmole) and there so very little A− in solution to start but the incoming strong base will drive the equilibrium to the right and produce an amount of mmoles of A− equivalent to the incoming strong base (4 mmole) so you will add that to the conjugate weak base (which was originally ~ 0)

[H+] = Ka

= 1.4 x 10-4

= 1.4 x 10-4

[H+] = 2.1 × 10−4 pH = 3.68

118. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 12.5 ml of 0.40 M KOH?

This is the halfway point of the titration, because half of the weak acid is reacted (10 mmol − 5 mmol), and half of the conjugate weak base forms (5 mmol)

[H+] = Ka

= 1.4 x 10-4

= 1.4 x 10-4

So always at the halfway point, the Ka = [H+] = 1.4 × 10−4 and pH = 3.85

119. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 20.0 ml of 0.40 M KOH?

At this point you are again only partway through the titration so use the buffer equation.

Calculate the millimoles of lactic acid (0.2 M × 50 ml) and millimoles of incoming strong base (0.4M × 20ml)

Subtract the incoming strong base from the weak acid, to find out the remaining un-neutralized acid (2 mmole) and there is so very little A− in solution to start but the incoming base will drive the equilibrium to the right and produce an amount of mmoles of A− equivalent to the incoming strong base (8 mmole) so you will add that to the conjugate weak base (which was originally ~ 0)

[H+] = Ka

= 1.4 x 10-4

= 1.4 x 10-4

[H+] = 3.5 × 10−5 M pH = 4.46

120. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 25.0 ml of 0.40 M KOH?

At this point you are at the equivalence point and all of the WA has been neutralized − the equilibrium has been driven all the way to the right, but the pH is NOT 7 because there will be conjugate WB, (A−) in the solution which does not just “sit there,” but hydrolyzes with water causing the pH to be above 7. So here you must switch gears

You have already calculated the millimoles of WA, and that will be the amount of conjugate WB, A− in solution because the incoming strong base has driven the neutralization “all the way” and all the WA has been converted to cjWB. Thus [A−] = 10 mmole divided by total volume (75 ml) = 0.133 M

You can then use the [A−] in a simple “x2” WB calculation to determine the x = [OH−]. Of course you need to calculate the Kb from the Ka remember: Ka × Kb = 1 × 10−14 thus Kb = 7.1×10−11 Kb = [OH−] [HA] = x2 x = [OH−] = 3.1 × 10−6 M [A−] [0.133]

pOH = −log[3.1 × 10−6] = 5.51 14.00 – pOH = 8.49

121. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 30.0 ml of 0.40 M KOH?

At this point in the titration the amount of strong base added exceeds the amount of weak acid.

You know that the equivalence point occurs at 25 ml of KOH added, thus 30 ml of KOH is 5 ml beyond the equivalence point resulting in an extra 2 mmole of strong base (since 0.40 M × 5 ml extra base)

Divide this extra 2 mmole of strong base by total volume (80 ml) to get 0.025 M of strong base in excess. (The rest of the incoming strong base was neutralized by H+ put out by the dissociating WA.) This SB which contributes far more OH− than the conjugate weak base does, so you need not concern yourself with any minute “extra” OH− contributed by the WB.

pOH = −log[0.025] = 1.60 14.00 – pOH = 12.40

122. Make a sketch of the titration curve using the volumes and pH values calculated for the titration of 50 ml of 0.20 M lactic acid with 0.40 M KOH. Placing a point on the graph for problems 117 −121 and #’s 123 & 124 below.

As you construct this graph, you will get a much more accurate sense of the curve if you make two more calculations and place two more points on the graph.

123. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 24.0 ml of 0.40 M KOH?

At this point, just like the calculation that you made in #119, you are again only partway through the titration so use the buffer equation. Calculate the millimoles of lactic acid (0.2 M × 50 ml) and millimoles of incoming strong base (0.4M × 24ml). Subtract the incoming strong base from the weak acid, to find out the remaining un-neutralized weak acid (0.4 mmole). Realize that there so very little A− in solution to start but the incoming strong base will drive the equilibrium to the right and produce an amount of mmoles of A− equivalent to the incoming SB (9.6 mmole) so you will add that to the cjWB (which was originally ~ 0)

[H+] = Ka

= 1.4 x 10-4

= 1.4 x 10-4

[H+] = 5.8 × 10−6 M pH = 5.23

124. Calculate the pH of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 26.0 ml of 0.40 M KOH?

At this point in the titration the amount of SB added exceeds the amount of WA, just like in #121.

You know that the equivalence point occurs at 25 ml of KOH added, thus 26 ml of KOH is 1 ml beyond the equivalence point. From previous calculations, we know that the mmole of WA is (0.2 M × 50 ml) 10 mmoles, and the mmol of incoming SB is more, 10.4 mmoles (0.40 M × 26 ml SB)

Subtract to get 0.4 mmole of excess SB, then divide by total volume (76 ml) to get 0.0055 M of strong base in excess. (The rest of the incoming SB was neutralized by H+ put out by the dissociating WA.) This SB which contributes far more OH− than the conjugate weak base contributes, so you need not concern yourself with any minute “extra” OH− contributed by the WB.

pOH = −log(0.025) = 2.28 14.00 – pOH = 11.72

125. Calculate the pH of 20.0 ml of a 0.060 M solution of ammonia, NH3.

Solve for pH before any titration with a simple “x2” set-up. Remember that the WB really is 0.060 − x, however the − x is too small to worry about. The AP exam will not require that you solve a problem in which you would need to use the quadratic equation. Thus Kb = [OH−] [HA] = x2 x = [OH−] = 1.04 × 10−6 M [A−] [0.060]

pOH = −log[1.04 × 10−6] = 2.98 14.00 – pOH = 11.02

126. What volume of 0.040 M HI would be needed to neutralize this 20.0 ml of a 0.060 M solution of ammonia, NH3?

At the equivalence point, the moles of base = moles of acid, so use:

MbVb = MaVa ; 0.06 M × 20 ml = 0.04 M × Vb

Thus the vol needed to reach the equivalence point is 30. ml of 0.04 M HI

127. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 5.0 ml of 0.040 M hydroiodic acid, HI?

At this point you are partway through the titration, but not halfway, thus you are in the “land of buffers.”

Calculate the mmol of base, ammonia (0.06 M × 20 ml = 1.2 mmol)

mmol of incoming strong acid (0.04 M × 5 ml = 0.2 mmol)

[OH−] = Kb

= 1.8 x 10-5

= 1.8 x 10-5

[OH−] = 8.9 × 10−5 M pOH = 4.05 pH = 14.00 – pOH = 9.95

128. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 15.0 ml of 0.040 M hydroiodic acid, HI?

This is the halfway point of the titration, and at this point, the [WB] = [cjWA+] thus

Kb = [OH−] = 1.8 × 10−5, thus pOH = 4.74 pH = 14.00 – pOH = 9.26

129. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 20.0 ml of 0.040 M hydroiodic acid, HI?

Just like question # 127, you are still partway through the titration and in the “land of buffers.” Have no fear....you can do this.

As calculated before the mmol of base, ammonia (0.06 M × 20 ml = 1.2 mmol)

mmol of incoming SA (0.04 M × 20 ml = 0.8 mmol)

[OH−] = Kb

= 1.8 x 10-5

= 1.8 x 10-5

[OH−] = 9.0 × 10−6 M pOH = 5.05 pH = 14.00 – pOH = 8.95

130. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 30.0 ml of 0.040 M hydroiodic acid, HI?

At this point you are at the equivalence point and all of the WB has been neutralized, and the equilibrium has been driven all the way to the right leaving only the conjugate weak acid, NH4

+ aka WB+ in the solution.

You have already calculated the millimol of WB, (0.06 M × 20 ml = 1.2 mmol), and since the equilibrium has been driven completely to the right, that number now is the amount of conjugate NH4

+, WB+ in solution.

Calculate the concentration of this conjugate weak acid 1.2 mmole by dividing by total volume 50. ml 1.2 mmol H+

= 0.024M M SA 50ml TotalVol

You can then use the [cjWA+], an acid, in a simple “x2” WA calculation to determine the [H+] but we’ll need the Ka for the calculation at the equivalence point. Thus Ka = 5.6 ×10−10

Ka = [H+] [A−] = 5.6 ×10−10 = x2 x = [H+] = 3.7 × 10−6 M [HA] [0.024]

pH = −log[3.7 × 10−6 M] = 5.44

131. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 50.0 ml of 0.040 M hydroiodic acid, HI?

At this point in the titration the amount of acid added exceeds the amount of base. From the calculation in question #126, we know that there are 20 ml extra acid beyond the equivalence point.

(0.04 M × 20 ml) results in an extra 0.80 mmol of excess strong acid.

Divide this extra 0.08 mmol of acid by the total volume of 70. ml; 0.08 mmol H+

= 0.0114M M SA 70ml TotalVol

Which contributes far more H+ than any H+ contributed by the conjugate weak acid (so you can ignore the weak acid’s contribution).

pH = −log[0.014] = 1.94

Make a sketch of the titration curve using the volumes and pH values calculated for the titration of 20 ml of 0.060 M ammonia with 0.040 M hydroiodic acid.

132. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 25.0 ml of 0.040 M hydroiodic acid, HI?

Just like question # 127 and #129, you are still partway through the titration and in the “land of buffers.”

As calculated before the mmol of base, ammonia (0.06 M × 20ml = 1.2 mmol)

mmol of incoming strong acid (0.04 M × 25.0 ml = 1.0 mmol)

[OH−] = Kb

= 1.8 x 10-5

= 1.8 x 10-5

[OH−] = 3.6 × 10−6 M pOH = 5.44 pH = 14.00 – pOH = 8.56

133. Calculate the pH of 20.0 ml of 0.060 M ammonia, NH3 after it has been titrated with a total of 35.0 ml of 0.040 M hydroiodic acid, HI?

Just like the calculation for #131, at this point in the titration the amount of acid added exceeds the amount of base. From the calculation in question #126, we know that there are 5 ml extra strong acid beyond the equivalence point.

(0.040 M × 5.0 ml) results in an extra 0.20 mmol of strong acid.

Divide this extra 0.20 mmol of SB by the total volume of 55 ml;

0.20 mmol H+ = 0.036M M SA

55ml TotalVol

Which contributes far more H+ than any H+ contributed by the conjugate weak acid (so you can ignore the weak acid’s contribution).

pH =−log(0.036) = 2.44

134. Sketch the curve that represents the titration of 40.0 ml of 0.010 M HCl (in the flask) with 0.016 M NaOH. Choose a few key points to calculate; starting pH, volume to reach equivalence point, pH at 50 ml)

135. On the same graph, sketch the curve that represents the titration of 40.0 ml of 0.010 M HCl (in the flask) with 0.016 M Ba(OH)2 (Choose a few key points to calculate; starting pH, volume to reach equivalence point, pH at 50 ml)

•pH at start of both titrations is starting HCl pH = 2.00

•volume at equivalence point use MV=MV

(0.01 M)(40 ml) = (0.016 M)V V = 25 ml

•At equivalence point, (always true for SA/SB) pH = 7

•At 50 ml, 25 ml extra base (for first titration)

(0.016 M)(25 ml) = M(90 ml) M = 0.0044 pH = 11.65

•At 50 ml, 37.5 ml extra base (for second titration)

(0.016 M)(37.5 ml) = M(90 ml) M = 0.0067 pH = 11.82

136. Sketch the curve that represents the titration of 20.0 ml of 0.0020 M NaOH (in the flask) with 0.001 M HNO3 (Choose just a few key points to calculate; starting pH, volume to reach equivalence point, pH at 50 ml)

•pH at start of titration is starting NaOH pH = 12.30

•volume at equivalence point use MV=MV

(0.002 M)(20 ml) = (0.001 M)V V = 40 ml

•At equivalence point, (always true for SB/SA) pH = 7

•At 50 ml, 10 ml extra acid

(0.001 M)(10 ml) = M(70 ml) M = 0.00014 pH = 3.85

137. Sketch the curve that represents the titration of 40.0 ml of 0.010 M nitrous acid, HNO2 (in the flask) with 0.016 M NaOH. Ka of HNO2 = 4.5 × 10−4 (Calculate a few key points; starting pH, pH and volume at equivalence point, “halfway, after 50 ml)

138. Sketch the curve that represents the titration of 40.0 ml of 0.010 M hydrocyanic acid, HCN with 0.016 M NaOH. Ka of HOCl = 3.0 × 10−8 (Calculate a few key points; starting pH, pH and volume at equivalence point, “halfway, after 50 ml)

•pH at start is a WA “x2” problem. 4.5 ×10−10 = x2 x = [H+] = 2.1 × 10−3 M pH = 2.67

[0.024]

•pH halfway = pKa pH = 3.35

•volume at equivalence point use MV=MV

(0.01 M)(40 ml) = (0.016 M)V V = 25 ml

•pH at equivalence point is an “x2” problem of cjWB 2.2 ×10−11 = x2 x = [OH−] = 3.7 × 10−7 M pH = 7.75

[0.0062]

•At 50 ml, 25 ml extra base (same for both titrations)

(0.016 M)(25 ml) = M(90 ml) M = 0.0044 pH = 11.65

•For second titration; pH at start is a WA “x2” problem. 3 ×10−8 = x2 x = [H+] = 1.7 × 10−5 M pH = 4.76

[0.01]

•pH halfway = pKa pH = 7.52

•pH at equivalence point is an “x2” problem of cjWB 3.3 ×10−7 = x2 x = [OH−] = 4.5 × 10−5 M pH = 9.66

[0.0062]

139. Sketch the curve that represents the titration of 30.0 ml of 0.020 M ammonia, NH3 (in the flask) with 0.015 M HCl. Kb of NH3 = 1.8 × 10−4 (Calculate a few key points; starting pH, pH and volume at equivalence point, “halfway, after 50 ml)

140. Sketch the curve that represents the titration of 30.0 ml of 0.020 M pyridine, C5H5N (in the flask) with 0.015 M HNO3. Kb of C5H5N = 1.7 × 10−9 (Calculate a few key points; starting pH, pH and volume at equivalence point, “halfway, after 50 ml)

•pH at start is a WB “x2” problem. 1.8 ×10−3 = x2 x = [OH−] = 1.9 × 10−3 M pH = 11.28

[0.02]

•pOH halfway = pKb pH = 10.26

•volume at equivalence point use MV=MV

(0.02 M)(30 ml) = (0.015 M)V V = 40 ml

•pH at equivalence point is an “x2” problem of cjWA 5.6 ×10−10 = x2 x = [H+] = 2.2 × 10−6 M pH = 5.66

[0.0086]

•At 50 ml, 10 ml extra acid (same for both titrations)

(0.015 M)(10 ml) = M(80 ml) M = 0.0019 pH = 2.73

•Again, pH at start is a WB “x2” problem. 1.7 ×10−9 = x2 x = [OH−] = 5.8 × 10−6 M pH = 8.76

[0.02]

•pOH halfway = pKb pH = 5.23

•pH at equivalence point is an “x2” problem of cjWA 5.9 ×10−6 = x2 x = [H+] = 2.2 × 10−4 M pH = 3.65

[0.0086]

This is simply a SA that has been diluted, need to calculate it’s concentration.

141. Calculate the pH of a solution made by pouring 5.0 ml of 0.20 M HCl into 100.

ml of water.

This is simply a SA that has been diluted, need to calculate it’s concentration.

M ×V = M ×V (0.2M)(5ml ) = (M)(105ml ) M = 0.0095M pH = 2.02

142. Calculate the pH of a solution made by pouring 10.0 ml of 0.20 M NaOH into 100. ml of water.

This is simply a SB that has been diluted, need to calculate it’s concentration.

M ×V = M ×V (0.2M)(10ml ) = (M)(110ml ) M = 0.018M pOH = 1.74

pH = 12.26

143. Calculate the pH of 100. ml of 0.20 M nitrous acid, HNO2 (Ka = 4.5 × 10−4).

This is a WA, nothing done to it. This is an “x2” problem. 4.5 ×10−4 = x2 x = [H+] = 0.0095 M pH = 2.02

[0.20]

144. Calculate the pH of 100. ml of 0.20 M ammonia, NH3 (Kb = 1.8 × 10−5).

This is a WB, nothing done to it. This is an “x2” problem. 1.8 ×10−5 = x2 x = [OH−] = 0.0019 M pOH = 2.02

[0.20] pH = 11.28

145. Calculate the pH of a solution made by dissolving 1.38 g of NaNO2 (MM = 69.0 g/mol) into enough water to produce 100. ml of solution.

NaNO2 is a soluble alkali salt that puts a weak base, NO2− ion into solution,

nothing else has been done to the weak base ion, thus, this is an “x2” problem

Ka × Kb = Kw 4.5 ×10−4 × Kb = 1×10−14 Kb = 2.2 ×10−11

1.38 g NaNO2 1 mol NaNO2 = 0.20 M 0.1 L 69.0 g NaNO2

2.2 ×10−11 = x2 x = [OH−] = 2.1 ×10−6M pOH = 5.68 [0.20] pH = 8.32

146. Calculate the resulting pH after adding 10.0 ml of 0.20 M NaOH into the solution from problem # 143.

This is a WA with incoming SB, not at halfway, not yet at the equivalence point.

M ×V = mol (0.2M)(100ml ) = 20mmolWA (0.2M)(10ml ) = 2mmol incomingSB

[H+] = Ka

= 4.5 x 10-4

= 4.5 x 10-4

[H+] = 0.00405 M pH = 2.39

147. Calculate the resulting pH after adding 20.0 ml of 0.20 M HCl into the solution from problem # 144.

This is a WB with incoming SA, not at halfway, not yet at equivalence point.

[OH−] = Kb

= 1.8 x 10-5

= 1.8 x 10-5

[OH−] = 7.2 × 10−5 M pOH = 4.14 pH = 14.00 – pOH = 9.86

148. Calculate the pH of a solution produced by dissolving 1.38 g of NaNO2 into 100. ml of 0.20 M nitrous acid, HNO2. (Assume that there is no volume change.)

This is a cjWB with its WA = buffer but nothing done to it − no incoming SA or SB. You can calculate [OH−] or [H+] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in #145. I will not do that extra work…

mmol cjWB = 1.38g ÷ 69 g/mol = 0.02mol = 20mmolcjWB

0.2M ×100 = 20mmolWA

[H+] = Ka

= 4.5 x 10-4

[H+] = 4.5 x 10-4 M pH = 3.35

149. Calculate the resulting pH after adding 10.0 ml of 0.20 M NaOH into the solution from problem # 148.

This is a cjWB with its WA = buffer but this time with incoming SB. You can calculate [OH−] or [H+] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in #145. I will not…

mmol cjWB = 1.38g ÷ 69 g/mol = 0.02mol = 20mmolcjWB

0.2M ×100 = 20mmolWA 0.2M ×10 = 2mmolSB

[H+] = Ka

= 4.5 x 10-4

= 4.5 x 10-4

[H+] = 3.68 x 10-4 M pH = 3.439

150. Calculate the resulting pH after adding 20.0 ml of 0.20 M HCl into the solution from problem # 148.

This is a cjWB with its WA = buffer but this time with incoming SA. You can calculate [OH−] or [H+] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in #145. I will do that this time to show you the difference…but the results will be the same.

Ka × Kb = Kw 4.5 ×10−4 × Kb = 1×10−14 Kb = 2.2 ×10−11

mmolcjWB = 1.38g ÷ 69 g/mol = 0.02mol = 20mmolcjWB

0.2M ×100 = 20mmolWA 0.2M × 20 = 4mmolSA

[OH−] = Kb

= 1.8 x 10-5

= 1.8 x 10-5

[OH−] = 1.5 × 10−11 M pOH = 10.83 pH = 14.00 – pOH = 3.17

151. Calculate the resulting pH after adding 110. ml of 0.20 M HCl into the solution from problem # 148.

This is a cjWB with its WA = buffer but this time with incoming SA, in which we have gone 2 mmol beyond the buffer capacity.

Anytime we go beyond the buffer capacity, or beyond an equivalence point. It is only about the excess incoming strong (acid or base) and total volume to calculate the molarity of the excess.

mmolcjWB = 1.38g ÷ 69 g/mol = 0.02mol = 20mmolcjWB

0.2M ×100 = 20mmolWA 0.2M ×110 = 22mmolSA

22mmol – 20mmol = 2mmol H+

[H+] = 2mmol = 0.0095M pH = 2.02 210ml

152. Calculate the resulting pH after adding 120. ml of 0.20 M NaOH into the solution from problem # 148.

This is a cjWB with its WA = buffer but this time with incoming SB, in which we have gone 4 mmol beyond the buffer capacity. Anytime we go beyond the buffer capacity, or beyond an equivalence point. It is only about the excess incoming strong (acid or base) and total volume to calculate the molarity of the excess.

mmolcjWB = 1.38g ÷ 69 g/mol = 0.02mol = 20mmolcjWB

0.2M ×100 = 20mmolWA 0.2M ×120 = 24mmolSB

[OH−] = 4 mmol = 0.018M pOH = 1.74 pH = 12.26 220ml

153. Calculate the resulting pH after adding 100. ml of 0.20 M HCl into the solution from problem # 148

WB with incoming SA, and we’ve reached the equivalence point ! Now it is an x2

problem for the cjWB swimming in the total volume.

0.2M ×100 = 20mmolWB 0.2M ×100 = 20mmol incomingSA

Ka × Kb = Kw Ka ×1.8 ×10−5 = 1×10−14 Kb = 5.6 ×10−10

[cjWB] = 20 mmol cjWA = 0.10M 200ml

5.6 ×10−10 = x2 x = [H+] = 7.5 ×10−6 M pH = 5.12 [0.10]

154. Calculate the resulting pH after adding 100. ml of 0.20 M NaOH into the solution from problem # 143.

WA with incoming SB, we’ve reached the equivalence point! Now it is an x2 problem for the cjWA swimming in the total volume.

0.2M ×100 = 20mmolWA 0.2M ×100 = 20mmol incomingSB

Ka × Kb = Kw Kb ×4.5 ×10−4 = 1×10−14 Kb = 2.2 ×10−11

[cjWB] = 20 mmol cjWA = 0.10M 200ml

2.2 ×10−11 = x2 x = [OH−] = 1.5 ×10−6M pOH = 5.83 [0.10] pH = 8.17

155. Calculate the resulting pH after adding 100.0 ml of 0.22 M NaOH into the solution from problem # 143.

This is a WA with incoming SB, in which we have gone beyond 2 mmol beyond the equivalence point. Anytime we go beyond the equivalence point, it is only about the excess incoming SB and total volume to calculate the molarity of the excess.

0.2M ×100 = 20mmolWA 0.22M ×100 = 22mmolSB.

[OH−] = 2mmol = 0.010M pOH = 2.00 pH = 12.00 200ml

156. Calculate the resulting pH after adding 100.0 ml of 0.22 M HCl into the solution from problem # 144.

This is a WB with incoming SA, in which we have gone beyond 2 mmol beyond the equivalence point. Anytime we go beyond the equivalence point, it is only about the excess incoming SA and total volume to calculate the molarity of the excess.

0.2M ×100 = 20mmolWB 0.22M ×100 = 22mmolSA

[H+] = 2mmol = 0.010M pH = 2.00 210ml

157. Calculate the resulting pH after adding 10.0 ml of 0.20 M HCl into the solution from problem # 143.

WB with SB added to it. This is a curveball, AP is not likely to ask this question, WA with SA added to it. This is a curveball, AP is not likely to ask this question, but they have once before...so just in case. The WA is unimportant compared to the amount of SA coming in, while it is tempting to add the mmol of WA present from problem # 143 (0.95 mmol), the incoming acid would actually suppress the amount of H+ contributed by the WA; think LeChatelier’s principle.

[H+] = 2mmol SA = 0.018M pH = 1.74 110 totalvol

158. Calculate the resulting pH after adding 20.0 ml of 0.20 M NaOH into the solution from problem # 144

WB with SB added to it. This is a curveball, AP is not likely to ask this question, but they have once before...so just in case. The WB is unimportant compared to the amount of SB coming in, while it is tempting to add the mmol of WB present from problem # 144 (0.19 mmol), the incoming base would suppress the amount of OH− contributed by the WB; think LeChatelier’s principle.

[OH−] = 4mmol = 0.033M pOH = 1.74 pH = 12.52 120ml

159. For (a) − (g) which of the following salts, when dissolved in water, at 25ºC, cause a change in pH?

If there is a pH change, will the pH be above or below 7? Write a hydrolysis reaction that illustrates the cause of the pH change. Calculate the pH if 2.0 g of the salt is dissolved in 100 ml of water.

a. barium chloride, BaCl2 (208.2 g/mol)

•pH will remain at 7.

•BaCl2 → Ba2+ + 2 Cl− (note the single arrow since BaCl2 is a soluble ionic compound).

•The Ba2+ ions are “pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions. The Cl− ions are also “pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions.

b. ammonium perchlorate, NH4ClO4 (117.492 g/mol)

•pH will change to below 7, acidic.

•NH4ClO4 → NH4+ + ClO4

− (note the single arrow since NH4NO3 is a soluble ionic compound)

•The ClO4− ions are are “pathetic” meaning they do not hydrolyze in water and

thus do not produce any extra H+ or OH+ ions. However, the ammonium ions are a WA, thus the NH4

+ ions partially dissociate in water and produce H+ ions.

•NH4+ ⇌ H+ + NH3

2 g ÷ 117.492 g/mol = 0.017mol ÷ 0.1 L = 0.17 M

Ka × Kb = Kw Ka ×1.8 ×10−5 = 1×10−14 Ka = 5.6 ×10−10

5.6 ×10−10 = x2 x = [H+] = 9.7 ×10−6M pH = 5.01 [0.17]

c. sodium cyanide, NaCN (49.01 g/mol)

•pH will change to above 7, basic.

•NaCN → Na+ + CN− (note the single arrow since NaCN is a soluble ionic compound)

•The Na+ ions are “pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions. But the CN− ions are a conjugate WB which hydrolyzes in water and produces extra OH− ions.

•CN− + H2O ⇌ OH− + HCN

2 g ÷ 49.01 g/mol = 0.0408 mol ÷ 0.1 L = 0.408 M

Ka × Kb = Kw Kb ×3.5 ×10−4 = 1×10−14 Kb = 2.9 ×10−11

2.9 ×10−11= x2 x = [OH−] = 3.4 ×10−6M pOH = 5.47 [0.408] pH = 8.53

d. ammonium nitrate, NH4NO3 (80.052 g/mol)

•pH will change to below 7, acidic.

•NH4NO3 → NH4+ + NO3

− (note the single arrow since NH4NO3 is a soluble ionic compound)

•The NO3− ions are “pathetic” meaning they do not hydrolyze in water and thus

do not produce any extra H+ or OH− ions. But the NH4+ ions are a conjugate

WA and partially dissociate in water and produce H+ ions.

•NH4+ ⇌H+ + NH3

2 g ÷ 80.052 g/mol = 0.025 mol ÷ 0.1 L = 0.250 M

Ka × Kb = Kw Ka ×1.8 ×10−5 = 1×10−14 Ka = 5.6 ×10−10

5.6 ×10−10 = x2 x = [H+] = 1.2 ×10−5M pH = 4.93 [0.250]

e. potassium acetate, KC2H3O2 (98.144 g/mol)

•pH will change to above 7, basic.

•KC2H3O2 → K+ + C2H3O2− (note the single arrow since KC2H3O2 is a soluble

ionic compound) The K+ ions are “pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions. But the C2H3O2

−

ions are a conjugate WB which hydrolyzes in water and produces extra OH− ions.

•C2H3O2− + H2O ⇌ OH− + HC2H3O2

2 g ÷ 98.144 g/mol = 0.0204 mol ÷ 0.1 L = 0.204 M

Ka × Kb = Kw Kb ×3.5 ×10−4 = 1×10−14 Kb = 5.6 ×10−10

5.6 ×10−10= x2 x = [OH−] = 1.1 ×10−5M pOH = 4.97 [0.204] pH = 9.03

f. diethylamide bromide, (C2H5)2NH2Br (154.046 g/mol)

•pH will change to below 7, acidic.

•(C2H5)2NH2Br → (C2H5)2NH2+ + Br− (note the single arrow since (C2H5)2NH2Br

is a soluble ionic compound, you should be able to recognize this because of the Br− at the end of the molecule and the acid/ base context of the question)

•The Br− ions are “pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions. But the (C2H5)2NH2

+ ions partially dissociate in water and produce H+ ions.

• (C2H5)2NH2+ ⇌ H+ + (C2H5)2NH

2 g ÷ 154.06 g/mol = 0.013mol ÷ 0.1 L = 0.13 M

Ka × Kb = Kw Ka ×1.3 ×10−3 = 1×10−14 Ka = 7.69 ×10−12

7.69 ×10−12 = x2 x = [H+] = 9.99 ×10−7M pH = 6.00 [0.130]

g. aluminum perchlorate, Al(ClO4)3 (325.33 g/mol)

•pH will change to below 7, acidic.

•Al(ClO4)3 → Al3+ + ClO4− (note the single arrow since ClO4 is a soluble ionic

compound, you should be able to recognize this because of the ClO4− at the

end and the acid/base context)

•The Al3+ ions interact with water and produce H+ ions. But the ClO4− ions are

“pathetic” meaning they do not hydrolyze in water and thus do not produce any extra H+ or OH− ions.

•Al(H2O)63+ Al(H2O)5(OH)2+ + H+ (Note: You would probably never have to

write this equation.)

2 g ÷ 324.33 g/mol = 0.00615mol ÷ 0.1 L = 0.0615 M

1.5 ×10−3 = x2 x = [H+] = 9.6 ×10−4M pH = 3.02 [0.0615]

160. For (a) − (n) will the pH be 7, above 7, or be below 7. The trick to the following questions is to be on the lookout for the SA “pathetics,” negative ions of SA (Cl−, Br−, I−, NO3

−, SO42−, ClO4

−, ClO3−), and the SB pathetics,

Group 1 & 2 positive ions (Li+, Na+, K+, Rb+, C+, Ba2+, Sr2+, Ca2+). If you know who the pathetics are, then all other positive ions are conjugate weak acids, and (nearly) all other negative ions are conjugate weak bases.

a. equal volumes of 1 M HF and 1 M HCl

KCl is made of two pathetic ions, and HCl is a SA. Thus this is NOT a buffer and is has a pH below 7.

b. equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2

HF is a WA, and HCl is a SA, thus this is nothing more than lots of acid; NOT a buffer, just pH below 7.

c. equal volumes of 1 M HF and 0.5 M NaOH

HC2H3O2 is a WA, and for NaC2H3O2 , the Na+ is a pathetic and the C2H3O2− is

the conjugate WB of the WA, HC2H3O2. Since both the WA and cjWB are present in equal quantities, this is a buffer with pH below 7.

d. equal volumes of 1 M NH3 and 1 M NH4NO3

HF is a WA, and NaOH is a SB. Look at the molarities to see that the SB will neutralize half the WA, thus half of the WA will be left, and half of the WA, HF, will turn into cjWB, F−. This is an acidic buffer with pH below 7.

e. 50 ml of 1 M NH3 and 25 of 1 M HNO3

NH3 is a WB, and HNO3 is a SA. Since the SA will neutralize half the WB, half of the WB will be left, and half of the WB will turn into cjWA. This is a basic buffer with pH above 7.

f. 50 ml of 1 M NaOH and 25 of 1 M HNO3

NaOH is a SB and HNO3 is a SA. We have twice as much base as acid, but since there is no WA and no cjWB, this is NOT a buffer. It is simply more SB present than SA, thus the pH will be above 7.

g. 100 ml of 1 M NH3 and 50 ml of 1 M NaOH

NH3 is a WB, and NaOH is a SB, thus this is nothing more than lots of base; NOT a buffer, just pH above 7.

h. equal volumes of 1 M NaBr and 1 M KCl

NaBr is made of two pathetic ions, Na+ and Br−, and KCl is also made out of two pathetics, K+ and Cl−. Thus this is NOT a buffer just lots of non-hydrolyzing ions, thus the pH is 7.

i. equal volumes of 1 M KOH and 1 M HClO4

KOH is made of a SB, and HClO4 is a SA. Since the SA and SB are present in equal quantities, neutralization occurs (H2O is formed) and all that remains are non-hydrolyzing, pathetic ions (K+ and ClO4

−) thus the pH is 7.

j. equal volumes of 1 M CsOH and 1 M HF

CsOH is made of a SB, and HF is a WA. Since the SB and WA are present in equal quantities, neutralization occurs (H2O is formed) and what remains in the beaker is a pathetic, Cs+, and a cjWB, F−, thus the pH is above 7.

k. equal volumes of 1 M HC2H3O2 and 2 M NaOH

HC2H3O2 is made of a WA, and NaOH is a SB. Since the SB is present in twice the quantity needed to neutralize the WA, there is lots of excess SB left over, thus the solution will NOT be a buffer with a pH above 7.

l. equal volumes of 1 M CsF and 2 M HCl

CsF is made of a pathetic and a cjWB, and HCl is a SA. Since the cjWB and SA are present in equal quantities, neutralization occurs (H2O is formed) and what remains in the beaker is are two pathetics, Cs+ and Cl−, and a cjWA, HF. Thus the pH is below 7.

m. equal volumes of 1 M H3PO4 and 1 M NaOH

H3PO4 is made of a WA, and NaOH is a SB. Since enough SB is present to neutralize one of the ionizable H+ ions, H2PO4

−, will remain in the beaker. This is a WA, with very little of it’s conjugate WB present and is thus NOT a buffer, but will be acidic with a pH below 7.

n. equal volumes of 1 M NaH2PO4 and 1 M Na2HPO4

NaH2PO4 is made of a pathetic, Na+, and a WA, H2PO4−. Na2HPO4 is a is a

pathetic, Na+, and a WB, HPO42− (the cjWB of H2PO4

−). This is a buffer, and would likely have a pH ~7ish, the pKa of the WA, H2PO4

−.

161. b The lead and aluminum nitrate salts produce LOTS of nitrate anions in solution. The substance that dissolves to give the least number of nitrites in solution would at first appear to be both a, and b since they have only one nitrate compared to c and d, however, all of a ionizes, but only some of b ionizes because it is a WA.

162. c To get a conjugate base, you must react the HSO4

− as an acid, and the remaining ion will be its conjugate base. HSO4

− ⇌ H+ + SO42−. Thus

the SO42− is the conjugate base.

163. a If pH 2 is −log(1 × 10−2) and pH 3 is −log(1 × 10−3) then it would stand to

reason that the pH of some concentration between 1 × 10−2 and 1 × 10−3 would give a pH in between 2 and 3. Since is [H+] = 0.0025 is between 1x10−2 and 1x10−3, the only appropriate pH option would be choice a

164. c First calc the pOH (14 – 5 = 9) Then “undo” the pOH of 9 to get c. (solve:

10−9 = [OH−]) 165. c First calc the pH (14 – 4.282 = 9.718) Since we know that pH = 9 when

[H+] = 1 × 10−9 and we know that pH = 10 when [H+] = 1 × 10−10 so we know that the actual [H+] must be less than 1 × 10−9 and greater than 1 × 10−10 thus the only option even close is c (If you had a calculator, which you don’t, it would be an easy solve: 10−9.718 = [H+])

166. e Since Ba(OH)2 dissolves to produce 2 OH− ions in solution, [OH−] is twice

the molarity given, 0.00010 which gives a pOH of 4, thus a pH of 10. 167. d “Undo” the pH (10−5.00) to get the [H+] = 1.0 × 10−5. The [H+] = [Z−]

because they are related to each other in 1:1 ratio. The equilibrium [HZ] is 0.020 – x (which is too small to worry about from a sig fig point of view). Then solve the Ka expression.

Ka = [H+][Z−] so Ka = [1.0 x 10-5] [1.0 x 10-5] [HZ] [0.020] 168. b Since [H+] [OH−] = 1 × 10−14 in room temperature pure water, we know

that [H+] = [OH−] = 1 × 10−7

169. c In a basic solution, [OH−] is greater than in water, while [H+] is less than in

water. 170. b Kw like all other Keq’s are temperature dependent and thus Kw is not

always 1.0 × 10−14 171. d Salts are basic (pH above 7) when the negative anion is the conjugate

base of a weak acid. Thus in K2CO3, the CO32-, the weak conjugate

base of H2CO3 and in NaF, the F− is the weak conjugate base of HF. Further, NH4Cl contains a conjugate weak acid, NH4

+ and Cu(NO3)2

contains the (Lewis) acid forming Cu2+ ion which hydrolyses one of the waters that solvates around it when in solution.

172. e Salts are neutral only if they contain the “pathetic” ions, i.e. anions ions

from strong acids (Cl−, Br−, I−, NO3−, SO4

2−, ClO4−, ClO3

−) and cations from strong bases (group I and II metal ions), thus none of the salts presented meet those criteria.

173. c It is important to notice that this is a salt that is pH and thus pOH

changing. The anion, NO2− is the conjugate weak base from the WA

HNO2. Thus this is an “x2” problem of a WB. First you must determine

the Kb. Since Ka × Kb = Kw, Kb =

then Kb = 2 × 10−11 then

solve 2.0 × 10−11 =

to get 2.0 × 10−7 will give a pOH in the 6-

ish range, option c 174. b Even without being given a Ka, you would know that it must be acidic,

thus pH = 1 or pH = 3 are the only reasonable options. But a 0.1 M SA would be [H+] = 0.1 for a pH of 1, and a WA must be less than this, thus b is the only reasonable option.

175. c This is an “x2” problem for a WA, to calculate [H+] and then convert to

percent dissociation which is, %dissociation =

So solve 9 × 10−4 =

to get [H+] = 0.03 M which would be

%dissoc =

= 3 %

176. e First you must “unlog” to solve for [H+] = 1 × 10−4, then plug into the Ka

expression and solve Ka =

177. c This is an “x2” problem, when solving for x = [H+] from which you can

determine the pH. 1.0 × 10−10 =

= 1 × 10−6 which solves for a

pH of 6.00 178. a This also is an “x2” problem, when solving for x = [OH−].

4.0 × 10−10 =

x= 2 × 10−5 M

179. d Since the K for the reaction is less than 1, we know that the equilibrium is

“more to the left.” This tells us that the acid and base on the left are more likely to be weaker than the acid and base on the right. You can tell which are acids and bases by remembering that an acid donates H+ and bases accept H+.

180. d Since this is a diprotic acid, when the complete ionization occurs, the

dissociation constant for the total ionization would be the product of each of the two dissociations. For H2A ⇌ H+ + HA− , Ka1 and for HA−

⇌ H+ + A2−, Ka2, thus the K for the overall equation would be Ka

181. d Remember, the bases are the proton acceptors. 182. b Since the K given is smaller than 1, the equilibrium position lies more on

the reactant than product side, thus the product base is stronger, and the reactant base is weaker. In addition, you can also conclude that the product acid is stronger, and the reactant acid is weaker, however, that was not an option.

183. a [H+] concentration of SA is simple because the complete ionization

means the molarity of the acid will equal the molarity of the [H+] but the incomplete ionization of acids means that these calculations require the use of the Ka and the equilibrium expression.

184. b The other conjugate acid base pair would be HOH and H3O+

185. b The conjugate of a SA is “pathetic” because it has is not able to hydrolyze

with water and change pH in any way. 186. c Compared to the quantity of H+ ions contributed by HF, although small

itself, is far more than the amount of H+ ions contributed by water.

187. c Percent is always part out of total, %dissociation =

188. d As you know if [OH−] = 1 × 10−5, then pOH = 5 and pH = 9 since pH +

pOH = 14 189. a In most cases, and certainly any that AP will present, the −x is always

small enough to be considered insignificant compared to the original molarity.

190. b Since %dissociation =

, 4% or 0.04 of 0.1 M will be [H+] = 0.004. Use this

in the equilibrium expression Ka =

and Ka =

then

solve. 191. e For SB, the [OH−] is the same as the molarity of the base since it

completely dissociates, thus −log [OH−], pOH would be 2 and pH = 12

192. b You can use the dilution equation for this. Ma Va = Mb Vb and do not forget that the base produces twice as much [OH−], thus (0.05 M)(Va) = 2(0.1 M)(50 ml) and solve.

193. b This is essentially the same question as #174. A 0.01 M SA would yield a

pH of 2, since hypobromous acid is weak, it will have a less acidic pH, but still lower than 7.

194. c Amphoteric means the ability to react both as an acid or a base. Any

diprotic weak acid that has lost one of its protons has the ability to lose a second, and act as an acid, or accept the first proton back, and act as a base.

195. d pH of 1 means [H+] = 0.1 and a pH of 2 means [H+] = 0.01, thus we can

work the dilution equation M1 V1 = M2 V2 so, (0.1 M)(1 L) = (0.01 M)(V) solve for V to get 10 L. Since there is already 1 L, you only need to add 9 L of water to bring the total volume to 10 L

196. a pH of 5 means [H+] = 1 x 10−5, and since HA ⇌ H+ + A−, the [H+] = [A−]

and since Ka =

substitute Ka =

and solve for Ka =

1 × 10−10 197. a Remember that SO4

2− is the conjugate base of the not so weak acid HSO4

− (we usually consider this to be a strong acid....) and since

Ka × Kb = Kw, Kb =

then solve.

198. d It is important to remember that sulfurous acid is a diprotic acid. Using

Ma Va = Mb Vb substitute 2(0.2 M)(100 ml) = (0.1 M)Vb and solve for 400 ml.

199. d At the equivalence point, there is water and salt in the solution, however

the soluble salt, Na2SO3 is a weak base. Specifically the sulfite ion is the conjugate weak base of the weak sulfurous acid.

200. c Consider the equilibrium reaction HC2H3O2 ⇌ H+ + C2H3O2

− and use

LeChatelier’s principle to realize that adding HCl, a source of H+ will cause the equilibrium position to shift to the left, but it will never be able to shift far enough to reduce the incoming [H+] to a value below what it started at.

201. c This question is a repeat of #194 but with a different diprotic acid. 202. d This is a strong acid. Thus 0.63 g/63 g/mol yields 0.01 mole/1L of ions,

this gives a pH of 2 203. a c, d, and e are weak acids, and HSO4

− is the second dissociation of H2SO4, thus HSO4

− is a weaker acid 204. b Since HC4H4O6

− is the conjugate base of H2C4H4O6 and Ka × Kb = Kw,

Kb =

then solve.

205. e HA ⇌ H+ + A−, the [H+] = [A−] and since Ka =

solving for [H+]

when [HA] = 1, is simply a square root. In the next 8 questions it is important to identify the two substances being combined.

You need to think about whether they are SA, SB, WA, WB. Remember to be on alert for salts that contain an ion that is a conjugate WB or conjugate WA. If you are combining a strong with a weak, be on alert for where you are in that combination: equivalence point, beyond equivalence point, or part way in the titration which is to say buffer.

206. d KCOOH is a salt in which COOH− is a conjugate weak base (the K+ is a

non-pH changing “pathetic”), the other particle, K2HPO3 is a salt that contains H2PO3

−, the conjugate weak base of the H3PO3 acid (the K+

is a non-pH changing“pathetic”). So, two weak bases do not make

a buffer and the pH would certainly be above 7.

207. c H3PO3 is a weak acid, and as stated in the previous problem, KH2PO3 contains H2PO3−, the conjugate weak base of the H3PO3 acid. Thus with an acid and it’s conjugate base in solution, you have a buffer with an acidic pH (You know that it is acidic because the Ka, 3 × 10−2 of the acid is larger than the Kb, 3.3 × 10−13 (since Ka × Kb = Kw) of the conjugate weak base.)

208. e CH3NH2 is a weak base and CH3NH3Cl is a salt which contains the weak

base’s conjugate weak acid CH3NH3+ (the Cl− is a non-pH changing

“pathetic”). So the weak base and it’s conjugate make a buffer with a pH above 7. (You know that it is basic because the Kb, 4.4 × 10−4 of the weak base is larger than the Ka, 2.3 × 10−11 (since Ka × Kb = Kw) of the conjugate weak acid.)

209. e This solution has a weak base, CH3NH2 and a strong acid, HCl, thus you

need to be on alert for the possible buffer formation. We can determine this by looking at the quantities of each substance. Equal volumes mean that the concentration is determining the quantities, and you can see that we have twice as much WB as the incoming SA. This means that you are at the halfway point of a titration, which means buffer. The pH will be above 7 because once again you know that it is basic because the Kb, 4.4 × 10−4 of the weak base is larger than the Ka, 2.3 × 10−11 (since Ka × Kb = Kw) of the conjugate weak acid.

210. d This solution is a di-protic weak acid, H3PO3 and a strong base, NaOH,

thus you need to be on alert for the possible buffer formation. We can determine this by looking at the quantities of each substance. Since we have a concentration that istwice the acid, both of the H+’s get ripped off and we are at the equivalence point which is certainly not a buffer and the pH would be above 7 because of the conjugate base that is in the water along with the pathetic Na+ ions from the SB.

211. b This solution contains a strong acid, HCl, and a strong base NaOH which

can never make a buffer. Thus you simple need to consider which substance is in excess. The concentrations are the same, but the HCl is present in double the volume. This means that you still have half of the strong acid remaining in solution causing a pH below 7

212. a NaCl is a salt made of two “pathetic” ions, thus it is not a pH changing

solution, and it is certainly not a buffer. 213. a This solution is made of a strong acid and a strong base. Equal volumes

and qual concentrations mean that we have added equal quantities, thus we are at the equivalence point, which will not be a buffer and will have a pH = 7.

214. b This solution may appear to be a buffer because it contains a weak acid,

HCOOH and a weak base, Na2HPO3 however, for a buffer to be a buffer, the WA and WB (HPO3

2−) must be conjugates of each other, which these two substances are not.

215. b Solution b contains a strong acid, HNO3, and a non-pH changing salt,

KNO3. This cannot be a buffer because there is no weak acid or weak base. The KNO3 has no effect on the strong acid, this means the pH will be very acidic.

216. e When equal quantities of strong acid and strong base are combined, the

pH will be 7 since you would be at the equivalence point. 217. c Remember that to have a buffer with a pH above 7, you need a weak

base and it’s conjugate weak acid. NH3 is a weak base and NH4NO3 is a salt which contains the conjugate weak acid, NH4

+. This is a buffer with a pH above 7. (You can assume this because you know that NH3 is a weak base, and if you looked up it’s Kb it would be greater than than the Ka of the conjugate weak acid.)

218. a For a buffer below 7 you need a weak acid and it’s conjugate weak base.

Oxalic acid is the weak acid, and the K is a salt which contains it’s conjugate weak base, HC2O4

−.

219. b Halfway to the equivalence point, in a titration, the WA = WB and thus the pH will = the [H+].

220. c The equivalence point on a graph is the middle of the steep vertical rise. 221. e As more base is added beyond the equivalence point, the excess base

will prevail. 222. b The term “balanced” buffer implies in the halfway-ish are of the titration.

223. d This is an “x2” problem. Thus 3.0 × 10−8 =

Look for the easy math.

x = [H+] = 3 × 10−5, This indicates a pH of 4-ish 224. b This is a buffer made of a weak base and its conjugate weak acid. We

can use the acid buffer formula [H+ ] = Ka

, but we will need the

Ka of the conjugate weak acid. We can calculate that since Ka × Kb = Kw, Ka = 1 × 10−6.

Next we need to calculate the quantities of both the WB and conjgate WA. (0.05 M)(0.10 L) = 0.005 moles WB and:

35g 1mol = 0.0050mol of chloride salte 69 g Which contains the conjugate weak acid. Then insert into the buffer

equation formula and of course halfway is very easy math.

[H+ ] = 1× 10−6

= 1 × 10-, thus pH = 6

225. e This obviously is a buffer problem, thus you can use the buffer formula.

[H+ ] = Ka

. Since we know the pH = 5, we know that

[H+]=1×10−5, and we can use the 10:1 ratio of WA to conjugate WB.

[1× 10−5 ] = Ka

then look for the easy math

226. d This problem is a weak acid that is being titrated with strong base. It may

be a buffer problem. You will know as soon as you calculate the quantities of each. WA (0.028 M)(100 ml) = 28 mmoles, the calculate incoming invader (0.36 M)(50 ml) = 18 mmoles, thus we have not

made it to the equivalence point. [H+ ] = Ka

. The amount of

incoming invader will shift the equilibrium: [H+] = 1.8 × 10−5

,

solve [H+] = 1.8 × 10−5

, and look for the easy math.

227. a When building a buffer, the most balanced, or ideal buffers have

approximately equal quantities of WA and WB. When this is true, [H+] = Ka, and pH of the buffer = pKa. Thus look for a weak acid with a pKa near the desired pH. Thus the second Ka of this acid, H2PO4

− has a pKa in the 7-ish range. So we need a salt of H2PO4− which we can find as , and we need the conjugate WB of this acid.

228. b Review your titration cuves for this answer. 229. a For this problem you need to look for a molecular acid, or look for a salt

that contains a conjugate weak acid ion. NH4Cl will satisfy that as NH4

+ is a conjugate weak acid (of NH3). Cl− is of course “pathetic”. “b and e” are neutral, and “c and d” contain conjugate weak base ions.

230. e Nitrite is a conjugate weak base of (nitrous acid), thus in water this

reaction occurs: NO2− + H2O ⇌ HNO2 + OH−

231. d To get a pH over 7, you need a conjugate weak base, thus you need to

look for a salt which contains an anion is not an anion of a strong acid. Thus any anion that is not Cl−, Br−, I−, NO3

−, SO42−

, ClO3−, and

ClO4−, will be an anion that is a weak base. Thus SO3

2− is the conjugate weak base of sulfurous acid, H2SO3, which is a weak acid. Note that NH4

+ is a conjugate weak acid of NH3, thus “a” would be acidic. “b” would be acidic because metal ions in solution (not groups 1 or 2) hydrolyze with the water (as a Lewis acid) to generate H+ ions, “c” has pH = 7, because K+ and are “pathetic” and “e” is of course simply acidic.

232. b In this problem nitrate ion will have no effect on any of the salts except

Pb(CN)2. Remember this is a Ksp situation. Consider this equilibrium: Pb(CN)2 ⇌ Pb2+ + CN− and when nitric acid is added, the H+ will

react with the CN− base and cause the Pb(CN)2 to dissolve more and shift right to generate more CN− to make up for what is being removed as HCN

233. e The fastest way to solve this problem is to cross out all the “pathetics”

and look for what ions are left and decide if the are weak acid ions or weak base ions. KNO3 and NaCl would be neutral, KClO would be basic because of the ClO− ion, and NH4NO3 would be acidic because of the NH4+ ion. Na2CO3 is basic because of the CO3

2− ion is a conjugate weak base of carbonic acid.

234. c Only c contains all “pathetics”. Items a, b, e would be acidic and item d

would be basic. 235. b When HNO2 is converted to HNO3, a strong acid − the oxidation number

of N changes from 3+ to 5+, oxidation 236. b NaHCO3, baking soda, in the presence of acid is a weak base. The

bicarbonate ion is a weak base and will be dissolved into Na+ and HCO3

−.