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Solution Equilibria I. Strong acids and Bases—pH

A. Self dissociation of water.

1. Recall that the following equilibrium takes place in aqueous solutions:

2H2O H3O+ + OH–

K = Kw = [H3O+][OH–] = 1.0x10–14 at room temperature.

2. This expression must be satisfied in all aqueous solutions, that is there is always

H3O+ and OH– ions present in these solutions.

Examples:

a. In a 0.15 M HCl solution;

[H3O+] = 0.15 M, [OH–] = 1.0x10–14

[H3O+]=1.0x10–14

0.15= 6.7x10

–14M

b. In a 0.054 M NaOH solution;

[OH–] = 5.4x10–2 M, [H3O+] = 1.0x10–14

[OH–]

=1.0x10–14

5.4x10–2=1.9x10

–13M

c. In a 0.15 M NaCl solution; [H3O+] = [OH–] = 1.0x10–7 M

B. pH 1. pH = -log[H3O+] where log = base 10 logarithm ∴ [H3O+] = 10

!

-pH a. If [H3O+] = 3.0x10-4, then pH = -log(3.0x10-4) = -(-3.52) = 3.52 b. If pH = 5.75, then [H3O+] = 10-5.75 = 1.78x10-6M 2. In general, p before any variable means - log(variable). pOH = - log[OH - ] pMg = - log[Mg2+] pK = - log K (K = the equilibrium constant) 3. Examples.

a. Consider the following solutions of some strong acids and bases.

Solution [H3O+] [OH- ] pH pOH pH+pOH 0.10M HCl 1.0x10-1 1.0x10-13 1.0 13.0 14.0 0.01M HCl 1.0x10-2 1.0x10-12 2.0 12.0 14.0 0.005M HCl 5.0x10-3 2.0x10-12 2.3 11.7 14.0 0.1M NaCl 1.0x10-7 1.0x10-7 7.0 7.0 14.0 0.02M NaOH 5.0x10-13 2.0x10-2 12.3 1.7 14.0 b. Note

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a. [H3O+][OH-] = 1.0x10-14 = Kw b. pH + pOH = 14.0 = pKw

c. An acidic solution = one in which pH < 7; [H3O+] > 1.0x10-7, [OH-] < 1.0x10-7. A neutral solution = one in which pH = 7; [H3O+] = [OH-] = 1.0x10-7. A basic solution = one in which pH > 7; [H3O+] < 1.0x10-7 , [OH-] > 1.0x10-7.

II. Weak acid/base equilibria. A. Equilibrium Constant Expressions. 1. Weak monoprotic acids. Consider acetic acid, HC2H3O2. It is a weak monoprotic acid. a. Equilibrium HC2H3O2 + H2O H3O+ + C2H3O2-

Ka = [H3O+][C2H3O2-]

[HC2H3O2] = 1.8x10-5 at 25 °C

Ka = acid dissociation constant b. Note. 1) [H2O] does not appear in the equilibrium constant expression. Since these are dilute aqueous solution reactions, the solvent H2O can be considered a "pure liquid" having a unit activity. 2) The equilibrium constant is very small. Therefore, at normal concentrations, only a very small percent of the acetic acid is dissociated, 2. Weak soluble bases. a. Consider ammonia, NH3. Due to the hydrolysis of NH3, an aqueous solution of NH3 in water will be basic. The weak base ‘ammonium hydroxide” is just an

aqueous solution of NH3. 1) Equilibrium NH3 + H2O NH4+ + OH-

Kb = [NH4+][OH-]

[NH3] = 1.8x10-5 at 25 °C

2) As was found for the acetic acid equilibrium, [H2O] does not appear in the equilibrium constant expression. 3) The reaction is called a hydrolysis reaction = reaction of a substance with water to give an acidic or basic solution where the H+ or OH- comes from the H2O.

b. The anions of the weak acids are the conjugate bases of the weak acids. Weak acids will yield fairly strong conjugate bases that can remove protons from water molecules, giving basic solutions. Therefore NaC2H3O2 will dissolve in water to give a basic solution. The C2H3O2- will hydrolyze to give a basic solution.

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1) Equilibrium C2H3O2- + H2O HC2H3O2 + OH-

Kb = [HC2H3O2][OH-]

[C2H3O2-] = 5.6x10-10 at 25 °C

3. Ka and Kb for conjugate acid base pairs. a. Self dissociation of water. 1) Equilibrium H2O + H2O H3O+ + OH- Kw = [H3O+][OH-] = 1.0x10-14 at 25 °C 2) This expression must be obeyed in all aqueous solutions. b. Consider the HC2H3O2 - C2H3O2- acid-base conjugate pair.

KaKb = [H3O+][C2H3O2-]

[HC2H3O2] x [HC2H3O2][OH-]

[C2H3O2-] = [H3O+][OH-] = Kw

That is, (1.8x10-5)x(5.6x10-10) = 1.0x10-14 The expression, KaKb = Kw, is a general one for any conjugate acid-base pair . 4. Example. Given that Kb for NH3 is 1.8x10-5, calculate Ka for NH4+.

Equilibrium NH4+ + H2O H3O+ + NH3

Ka = KwKb =

1.0x10-14

1.8x10-8 = 5.6x10-10 = [H3O+][NH3]

[NH4+]

B. Weak acid equilibrium calculations. 1. Given that Ka for HC2H3O2 = 1.8x10-5, calculate the percent dissociation of HC2H3O2 and the pH of a) a 1.0M HC2H3O2 solution and b) a 0.10M HC2H3O2 solution. a. 1.0M solution Let X = the decrease in the HC2H3O2 concentration required to reach equilibrium HC2H3O2 + H2O H3O+ + C2H3O2- Initial : 1.0M 1x10-7M 0 Equilibrium: 1.0 - X X X Have assumed that X >> 1x10-7 , that is, the [H3O+] is determined by the dissociation of the acetic acid.

Ka = [H3O+}[C2H3O2-]

[HC2H3O2[ = 1.8x10-5 = X2

1.0 - X

The value of X could be obtained by solving the quadratic equation,

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X2 + 1.8x10-5X - 1.8x10-5 = 0. However, the small value of Ka means that X will be a small number. Therefore, the assumption that 1.0 - X ≈ 1.0 is probability valid. Making this assumption simplifies the mathematics.

! 1.8x10-5

= 1.0

X2

or X = 1.8x10-5

= 4.24x10-3

[H3O+] = X = 4.24x10-3M ∴ pH = -log(4.24x10-3) = 2.37

% Dissociation = X

Initial Conc. (100) = 4.2x10-3

1.0 (100) = 0.42 %

Solving the quadratic equation gives an value of X = 4.23x10-3 and a pH = 2.37. Therefore, our simplifying assumption is valid. b. 0.10M solution Let X = the decrease in the HC2H3O2 concentration. HC2H3O2 + H2O H3O+ + C2H3O2- Initial: 0.10M 1x10-7 0 Equilibrium: 0.10 - X X X

Ka = 1.8x10-5 = X2

0.10 - X ≈ X2

0.10

! X = (0.10)(1.8x10-5

) = 1.34x10-3

[H3O+] = 1.34x10-3 pH = -log(1.34x10-3) = 2.87

% Dissociation = 1.34x10-3

0.10 (100) = 1.34%

If the quadratic equation is solved, X = 1.33x10-3 and pH = 2.88. Therefore, neglecting X is valid. 2. Consider the pH and percent dissociation as a HC2H3O2 solution is diluted. Quadratic X = Co(1.8x10-5) Co,M % pH % pH 1.0 0.42 2.37 0.42 2.37 0.1 1.33 2.88 1.34 2.87 0.01 4.15 3.38 4.24 3.37 1.0x10-3 12.55 3.90 13.42 3.87 5.0x10-4 17.26 4.06 18.97 4.02 1.0x10-4 34.37 4.46 42.43 4.37 5.0x10-5 44.64 4.65 60.00 4.52 1.0x10-5 71.55 5.14 134.2 4.87 1.0x10-6 94.99 6.02 1.0x10-7 99.45 7.00 1.0x10-8 100 7.00 a. Note that up to about 20% dissociation, X can be neglected in comparison to

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Co without introducing too much error. When the % dissociation is greater than 20% the quadratic should be solved. b. After the HC2H3O2 concentration falls below 1x10-7M the pH of the solution

remains at 7. The pH is now determined by H2O dissociation rather than acid

dissociation.

3. Polyprotic Acids. a. Consider H3PO4 . This is a triprotic acid. In an aqueous solution of H3PO4 the following equilibria exist.

H3PO4 + H2O H3O+ + H2PO4- Ka1 = [H3O+][H2PO4-]

[H3PO4] = 7.5x10-3

H2PO4- + H2O H3O+ + HPO42- Ka2 = [H3O+][HPO42-]

[H2PO4-] = 6.2x10-8

HPO42- + H2O H3O+ + PO43- Ka3 = [H3O+][PO43-]

[HPO42-] = 4.8x10-13

1) In any aqueous solution containing H3PO4, H2PO4-, HPO42-, or PO43- the above equilibria must be satisfied. 2) Note, there are a single species concentrations that must simultaneously satisfy all of the above equilibrium constant expressions. b. Example. Calculate the concentrations of all species in a 0.10M H3PO4 solution. since there are three dissociations, there are three unknowns that must be determined. Let X = the decrease in the [H3PO4] Let Y = the decrease in the [H2PO4-] Let Z = the decrease in the [HPO42-] The equilibrium concentrations are then: H3PO4 + H2O H3O+ + H2PO4-

0.01-X X+Y+Z X-Y H2PO4- + H2O H3O+ + HPO42- X-Y X+Y+Z Y-Z HPO42- + H2O H3O+ + PO43- Y-Z X+Y+Z Z These three unknowns can be substituted into the three equilibrium constant expressions and, in theory, solved. This system of equations is quite complicated but some

simplifying assumptions can be made. Note that the values of K decrease in the order Ka1 >> Ka2 >> Ka3 . This means that X >> Y >> Z. Therefore, the following assumptions are valid: [H3O+] = X+Y+Z ≈ X; [H2PO4-] = X-Y ≈ X; [HPO42-] = Y-Z ≈ Y.

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Ka1 = 7.5x10-3 = (X)(X)

0.10 - X ∴ X = 2.39x10-2 (the quadratic must be solved)

Ka2 = 6.2x10-8 = (X)(Y)

X ∴ Y = 6.2x10-8

Ka3 = 4.8x10-10 = (X)(Z)

Y ∴ Z = (Y)(4.8x10-10)

X = (4.8x10-10)(6.2x10-8)

2.39x10-3 = 1.3x10-18

∴ [H3O+] = [H2PO4-] = 2.39x10-2M, [HPO42-] = 6.2x10-8M, [PO43-] = 1.3x10-18M

C. Weak bases and buffer solutions.

1. Weak Bases. a. Calculate the pH of a 0.1M NH3 solution, given that Kb for NH3 = 1.8x10-5. Let Y = the decrease in the [NH3] required to reach equilibrium. NH3 + H2O NH4+ + OH- Initial: 0.1M 0 1x10-7M Equilibrium: 0.1 - Y Y Y Just as with acetic acid, we will assume that the H2O dissociation does not contribute to the pH.

Kb = [NH4+][OH-]

[NH3] = 1.8x10-5 = Y2

0.1 - Y ≈ Y20.1

Y = [OH-] = 1.34x10-3M pOH = -log(1.34x10-3) = 2.87 pH = 14.00 - 2.87 = 11.13 b. Hypocholorous acid, HOCl, is a weak acid whose Ka = 3.1x10-8. Calculate the pH of a 0.20M NaOCl solution. Let Y = the decrease in the OCl- concentration required to reach equilibrium. OCl- + H2O HOCl + OH- Initial: 0.20M 0 1x10-7M Equilibrium: 0.20 - Y Y Y

Kb = KwKa =

1.0x10-14

3.1x10-8 = 3.2x10-7 = [HOCl][OH-]

[OCl-]

3.2x10-7 = Y2

0.20-Y ≈ Y2

0.20 or Y = (0.20)(3.2x10-7) = 2.54x10-4

[OH-] = 2.54x10-4M pOH = 3.60 pH = 14.00 - 3.60 = 10.40

2. Solutions of weak acids + the salt of the weak acid. Buffer solutions.

a. Calculate the percent dissociation of HC2H3O2 and the pH of a solution that is

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1.0M in HC2H3O2 and 1.5M in NaC2H3O2.

Let X = the decrease in the [HC2H3O2] required to reach equilibrium. HC2H3O2 + H2O H3O+ + C2H3O2- Initial: 1.0M 1.0x10-7 1.5M Equilibrium: 1.0 - X X 1.5 + X Since NaC2H3O2 is a strong electrolyte, a 1.5M NaC2H3O2 solution is dissociated in water to give 1.5M Na+ and 1.5M C2H3O2-

Ka = 1.8x10-5 = X(1.5 + X)

1.0 - X ≈ 1.5X1.0 or X = 1.2x10-5

[H3O+] = 1.2x10-5M pH = -log(1.2x10-5) = 4.92

% Dissociation = 1.2x10-5

1.0 (100) = 0.0012 %

The solution is buffered at pH = 4.92. b. Buffer solution = solution of a weak acid and its conjugate base. It resists changes in pH. c. In the above example the [H3O+] is determined by Ka and the ratio of the acid concentration, CHA, and the salt (conjugate base) concentration, CS. In these solutions X (the amount of dissociation) will always be small compared to both CHA and CS. Therefore, the following relationships hold.

Ka = [H3O+]CS

CHA or [H3O+] = KaCHACS

Taking -log of both sides and rearranging one obtains

pH = pKa + log CS

CHA where pKa = -log(Ka) This equation is called the Henderson-Hasselbalch Equation and is useful in calculating the pH of buffers. Example: Use the Henderson-Hasselbalch Equation to calculate the pH of the 1.0M HC2H3O2 / 1.5M NaC2H3O2 buffer. pKa = -log(1.8x10-5) = 4.74

pH = 4.74 + log 1.51.0 = 4.74 + 0.18 = 4.92

d. Properties of a Buffer.

1) Since the pH depends on log CS

CHA , if the solution were diluted or concentrated

by the addition or removal of water, CS and CHA would be effected by the same

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amount and their ratio would not change. Therefore, the pH of the solution would not change. Care must be exercised in this statement. Remember that we should be using activities in these equation, not concentrations. On large dilution the activity coefficients of the acid and the salt may not change by the same amount and the pH will vary. 2) Additions of small amounts of foreign acids or bases will not affect the pH of

buffered solutions. Example: Consider a 500 ml solution of the 1.0M HC2H3O2 / 1.5M NaC2H3O2 buffer. Calculate the pH after 1 ml of 1M HCl or 1 ml of 1M NaOH is added. 1 ml 1M HCl. The HCl will react with the NaC2H3O2 to give HC2H3O2 and NaCl. mmol of HC2H3O2 originally present in the buffer = 500x1.0 = 500 mmol mmol of NaC2H3O2 originally present in the buffer = 500x1.5 = 750 mmol mmol HCl added = 1x1 = 1 mmol mmol HC2H3O2 present after addition = 500 + 1 = 501 mmol mmol NaC2H3O2 present after addition = 750 - 1 = 749 mmol

New pH = 4.74 + log 749/501501/501 = 4.74 + log 1.495 = 4.92

1 ml 1M NaOH The NaOH will react with the HC2H3O2 to give NaC2H3O2 and H2O. mmol HC2H3O2 present after addition = 500 - 1 = 499 mmol mmol NaC2H3O2 present after addition = 750 + 1 = 751 mmol

New pH = 4.74 + log 751/501499/501 = 4.74 + log 1.505 = 4.92

1 ml of 1M NaOH added to 500 ml of an unbuffered solution whose pH = 4.92. [H3O+] = 10-4.92 = 1.2x10-5M mmol of H3O+ originally present = (500)(1.2x10-5) = 6.0x10-3 mmol mmol of NaOH added = 1 mmol. ∴ have 1 - 6x10-3 ≈ 1 mmol OH- in excess in 501 ml. The [OH-] = 1/501 = 2.0x10-3M. pOH = =log (2.0x10-3) = 2.70 pH = 14.00 - 2.70 = 11.30

D. Titration and titration curve.

1. General. a. Titration = controlled addition from a buret, of a solution of one reactant , the titrant,

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to a solution of another reactant until an "end point" is reached. b. In an acid-base titration the end point is called the equivalence point. At the equivalence point one OH- ion has been added for each acidic proton, or vice versa. c. For the titration of monoprotic acids (HA) with monohydroxy bases (MOH), the

equivalence point is reached when VAMA = VBMB. Where VA = the volume of the acid of molarity MA, VB = the volume of the base of molarity MB. d. The titration curve is a plot of the pH of the solution versus the volume of titrant.

2. Suppose that 10.00 ml of a 0.20M HC2H3O2 solution is diluted to 100.00 ml with water and titrated with 0.01M NaOH. Calculate the pH after the following volumes of NaOH have been added: 0.00 ml; 10.00 ml; 50.00 ml; 100.00 ml; 150.00 ml; 199.00 ml; 200.00 ml; 201.00 ml; 210.00 ml; 220.00 ml.

Note, the equivalence point is a 200.00 ml of NaOH ((10.00)(0.20)0.01 ).

a. Initial pH Initial pH = pH of the 100 ml of HC2H3O2 before titration.

Dilution: [HC2H3O2] = (10.00 mL)(0.20M)

100.00 ml = 0.020M

Let X = the decrease in [HC2H3O2] HC2H3O2 + H2O H3O+ + C2H3O2- equilibrium: 0.020-X X X

Ka = 1.8x10-5 = [H3O+][C2H3O2-]

[HC2H3O2] = X2

0.020-X ≈ X2

0.020

X = (0.020)(1.8x10-5) = 6.0x10-4 = [H3O+]

pH = 3.22 ( Note, the % dissociation = 6.0x10-4

0.020 (100) = 3.0 % ∴ 0.02-X ≈ 0.02)

b. Buffer Region: All points from the initial pH to the equivalence point are in the buffer region of the titration curve. All solution in this region contain untitrated HC2H3O2 and NaC2H3O2 that is formed by the titration. These are buffer solutions. 1) 10.00 mL of 0.01M NaOH. mmol of HC2H3O2 originally present = (0.020M)(100.00 mL) = 2.00 mmol. mmol of NaOH added = (0.01M)(10.00 mL) = 0.10 mmol mmol of HC2H3O2 left untitrated = 1.90 mmol

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mmol NaC2H3O2 formed in the titration = 0.10 mmol The total volume = 110.00 mL.

∴ CHA = 1.90 mmol110.00 mL , CS =

0.10 mmol110.00 mL

The Henderson-Hasselbalch equation is pH = 4.74 + log( CS

CHA ).

∴ pH = 4.74 + log 0.10/110.001.90/110.00 = 4.74 - 1.28 = 3.46

Without the Henderson-Hasselbalch equation: HC2H3O2 + H2O H3O+ + C2H3O

!

2

" CHA- x x Cs+x if x is small CHA H3O+ Cs

!

KA =[H3O

+](CS)

(CHA)"[H3O] = KA

CHA

CS

=1.8x10#5 1.90 /110

0.10 /110 = 3.42x10-4

pH = 3.46 Note in the buffer region it is not necessary to determine the numerical values of CHA

or CS. The volumes cancel out. 2) 50.00 mL of 0.01M NaOH. mmol HC2H3O2 to be reacted = 2.00 mmol mmol of NaOH added = (0.01M)(50.00 mL) = 0.50 mmol. mmol HC2H3O2 left = 1.50 mmol mmol NaC2H3O2 formed = 0.50 mmol. The total volume = 150.00 mL

∴ CHA = 1.50 mmol150.00 mL , CS =

0.50 mmol150.00 mL

pH = 4.74 + log 0.50/150.001.50/150.00 = 4.74 - 0.48 = 4.26

3) 100.00 mL of 0.01M NaOH. mmol NaOH added = (0.01M)(100.00 mL) = 1.00 mmol mmol HC2H3O2 left = 2.00 - 1.00 = 1.00 mmol, mmol NaC2H3O2 formed = 1.00 mmol Half-equivalence point, have titrated half the acid. Note that at the half equivalence point CHA = CS . ∴ pH = 4.74 = pKa

4) 150.00 mL of 0.01M NaOH. mmol of NaOH added = (0.01M)(150.00 mL) = 1.50 mmol.

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mmol HC2H3O2 left = 2.00 – 1.50 = 0.50 mmol, mmol NaC2H3O2 formed = 1.50 mmol in 250 mL.

pH = 4.74 + log 1.50/250.00 0.50/250.00 = 4.74 + 0.48 = 5.22

5) 199.00 mL of 0.01M NaOH. mmol of NaOH added = (0.01M)(199.00 mL) = 1.99 mmol mmol HC2H3O2 left = 2.00 – 1.99 = 0.01 mmol, mmol NaC2H3O2 formed = 1.99 mmol in 299 mL.

pH = 4.74 + log 1.99/2990.01/299 = 4.74 + 2.30 = 7.04

c. Equivalence point 200.00 mL of 0.01M NaOH. mmol NaOH added = (0.01M)(200.00 mL) = 2.00 mmol = mmol of HC2H3O2 .

∴ have 2.00 mmol of NaC2H3O2 in 300.00 mL or a 2.00300 = 6.7x10-3 M NaC2H3O2

solution. The pH will be determined by the hydrolysis of the base NaC2H3O2. Let Y = decrease in the [NaC2H3O2] due to hydrolysis. C2H3O2- + H2O HC2H3O2 + OH- equilibrium: 6.7x10-3 - Y Y Y

Kb = KwKa =

1.0x10-14

1.8x10-5 = 5.6x10-10 = [HC2H3O2][OH-]

[C2H3O2-]

5.6x10-10 = Y2

6.7x10-3- Y ≈ Y2

6.7x10-3

∴ Y = [OH-] = (5.6x10-10)(6.7x10-3) = 1.93x10-6 pOH = 5.71 pH = 14.00 - 5.71 = 8.2 d. Post equivalence region. pH is determined by the excess titrant (NaOH). 1) 201.00 mL of NaOH. mmol of NaOH added = (0.01M)(201.00 mL) = 2.01 mmol mmol of OH- in excess = 2.01 mmol NaOH - 2.00 mmol HC2H3O2 = 0.01 mmol

[OH-]excess = 0.01 mmol

301 mL = 3.32x10-5M = [OH-]total

Any additional OH- obtained from either the hydrolysis of NaC2H3O2 or H2O is negligible. pOH = -log(3.32x10-5) = 4.48 pH = 14.00 - 4.48 = 9.52 2) 220.00 mL of NaOH. mmol of NaOH added = 2.20, mmol OH- in excess = 2.20-2.00 = 0.20 in 320 mL.

[OH-] = 0.20320 = 6.25x10-4 M. pOH = 3.20 pH = 14.00 - 3.20 = 10.80

3. Titration Curve.

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a. Plot of pH Vs Ml of NaOH. 1) The ‘S’ shaped curve is characteristic of the titration of an acid with a base.

2) The sharp increase in pH in Region C denotes the equivalence point. The pH of the equivalence is the pH at the midpoint of the break. In this example the equivalence point is at pH = 8.29. 3) The curve is composed of four distinct regions, labeled A, B, C, and D. Each will be discussed separately. b. Initial Point, point A. This is the pH of the acid before titration. If the acid is diluted, the dilution factor must be taken into account in calculating the initial pH. c. Buffer Region, region B. 1) Since the solution is buffered, the pH changes slowly with addition of titrant. 2) The pH at the half equivalence point (midway in the buffer region) = pKa. d. Equivalence Point Region, region C. 1) At the equivalence point one mmol of OH- has been added for every mmol of acid present. Therefore, the acid has been neutralized. However, due to the hydrolysis of the titration products, the pH at the equivalence point will not necessarily be equal 7 a) For titrations of strong acids + strong bases, at the equivalence point pH = 7. Example: HCl + NaOH ------> NaCl + H2O

2

4

6

8

10

12

pH

0 50 100 150 200 250ml 0.01M NaOH

A

B

C

D

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b) For titrations of weak acids + strong bases, at the equivalence point pH > 7 Example: HC2H3O2 + NaOH -----> NaC2H3O2 + H2O c) For titrations of strong acids + weak bases, at the equivalence point pH < 7. Example: NH3 + HCl ------> NH4Cl 2) The end point in a titration can be detected by using an indicator. Indicator is a weak acid, or base, whose acid form has a different color from the base form. Let HIn = the acid form of the indicator and In- = its base form. HIn + H2O H3O+ + In- Color A Color B If KIn = the equilibrium constant for this dissociation, then

KIn = [H3O+][In-]

[HIn]

pH = pKin + log (CIn-CHIn )

Since only a few drops of the indicator is used, the pH is determined by the position on the titration curve, which in turn fixes the ratio of CIn- to CHIn.

- The most sensitive color change would occur when the CIn-CHIn ratio changes

from about 1

10 to 101 .

- Therefore, one picks a indicator whose pKIn is approximately equal to the pH at the equivalence point. - Some indicators and their useful pH ranges are listed below. Acid Base Color change Indicator Color Color pH Range Thymol Blue red blue 1.2 - 2.8 Methyl Red red yellow 4.2 - 6.4 Bromothymol Blue yellow blue 6.0 - 7.6 Phenolphthalein colorless red 8.3 - 10.0 e. Post Equivalence Point Region, region D. The pH is determined solely by the excess titrant. 4. Titration curve of a polyprotic acid. a. In a titration, each proton is titrated progressively. That is the first proton before the second, the second before the third, etc. b. Consider a diprotic acid of the form H2A. When this is titrated the following reactions

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take place. First: 1) H2A + OH- ----> HA- + H2O When all the H2A is consumed: 2) HA- + OH- -----> A2- + H2O The titration curve will just be the curve for reaction 1) followed by that for reaction 2), stacked on top of one another. c. The following is the titration curve for the titration of a 50 ml sample of the acid H2A with 0.4M NaOH.

50 100 150 200 250 300 350 400

10

12

14

mL 0.4M NaOH

pH

8

6

4

2

0

0

2.25

6.10

1) The first equivalence point occurs after 100 ml of NaOH has been added. At that point the solution will contain only HA-, as its sodium salt. The concentration of H2A can be given by

M H

2A = (100)(0.4)

50 = 0.8 M

2) After the first 100 ml, the second proton, that of HA-, is titrated. Note that it takes an additional 100 ml to reach the second equivalence point.

3) the Ka's for the acids can be obtained from the two half equivalence points. At 50 ml the pH = 2.25 = pKa1 ∴ Ka1 = 10-2.25 = 5.6x10-3 At 150 ml the pH = 6.10 = pKa2 ∴ Ka2 = 10-6.10 = 7.9x10-7 III. Sparingly Soluble Salts. A. General. 1. The substance PbCl2 is a sparingly soluble salt. In a saturated solution of PbCl2 in water the following equilibrium takes place. PbCl2(s) Pb2+ + 2Cl- 2. The equilibrium constant expression is K = [Pb2+][Cl-]2 = KSP = Solubility Product Constant At 25 °C KSP for PbCl2 = 2.4x10-4. Note that the equilibrium is a heterogeneous one

15

and that the activity of the solid PbCl2 is equal to 1 and therefore does not appear in either the equilibrium constant expression or the reaction quotient, Q.

3. PbCl2 will dissolve until Q = [Pb2+][Cl-]2 = 2.4x10-4 2.4x10-4 is the maximum value for Q. If one tried to mix a solution in which

Q>2.4x10-4, a precipitate of PbCl2 would form. 4. Methods of obtaining KSP's. a. From measurements of the molar solubilities. Useful when KSP is not too small. 1) Molar solubility (S) = moles of a solute that will dissolve per liter of solution = The molarity of the salt in a saturated solution. 2) The solubility of PbCl2 (FM = 278.1) was found to be 1.09 g / 100 ml. Calculate KSP. Solubility in g / L = 1.09 g / 100 ml x10 = 10.9g/L.

Molar solubility = 10.9 g/L

278.1 g/mol = 3.92x10-2 mol/L.

Solution reaction: PbCl2(s) Pb2+ + 2Cl- ∴ [Pb2+] = 3.92x10-2 M; [Cl-] = 2(3.92x10-2 M) = 7.84x10-2 M KSP = [Pb2+][Cl-]2 = (3.92x10-2)(7.84x10-2)2 = 2.4x10-4

b. Many salts are so insoluble that the solubilities can not be measured directly. For example, KSP for HgS = 4.0x10-54. This means that in 833 L of a saturated solution of HgS there would be one Hg2+ ion and one S2- ion. Indirect, electrochemical, methods must be used to determine these KSP's. B. Uses of KSP. 1. Calculate molar solubilities in various solutions. Calculate the molar solubility of CaF2 in a) water, b) 0.01M Ca(NO3)2 , c) 0.01M NaF. Given KSP for CaF2 = 1.7x10-10.

Let S = the molar solubility a. In water CaF2(s) Ca2+ + 2F- Molarity in a saturated solution: S 2S

KSP = [Ca2+][F-]2 = (S)(2S)2 = 4S3

! S = 4

KSP3

= 4

1.7x10-10

3 = 3.5x10

-4 mol / L

16

In the saturated solution [Ca2+] = S = 3.5x10-4M, [F-] = 2S = 7.0x10-4M. b. In 0.01M Ca(NO3)2

CaF2(s) Ca2+ + 2F- Molarity in a saturated solution: 0.01 + S 2S Ca(NO3)2 is a strong electrolyte ∴ 0.01M Ca(NO3)2 = 0.01M Ca2+ + 0.02M NO3-

KSP = 1.7x10-10 = (0.01 + S)(2S)2 ≈ (0.01)(2S)2 = 0.04S2

∴ S = 1.7x10-10

0.04 = 6.5x10-5 mol / L

In the saturated solution [Ca2+] = 0.01M, [F-] = 1.3x10-4 M. c. In 0.01M NaF CaF2(s) Ca2+ + 2F- Molarity in a saturated solution: S 0.01+ 2S

KSP = (S)(0.01+2S)2 ≈ (S)(0.01)2 = (0.0001)(S)

S = 1.7x10-10

1.0x10-4 = 1.7x10-6 mol / L

In the saturated solution [Ca2+] = 1.7x10-6M, [F-] = 0.01 d. Note that the presence of either Ca2+ or F- in solution suppresses the solubility

of CaF2. This is called the common ion effect and would be expected form LeChatelier's principle. Since F- enters into the reaction quotient as a square, it is much more effective in suppressing the solubility of CaF2 than is Ca2+. C. Precipitation. 1. Prediction of precipitate formation. a. In any solution of CaF2, the reaction quotient, Q is given by [Ca2+][F-]2. In a saturated solution (at equilibrium) Q = KSP. If one tried to mix solutions so that, in the final solution, Q > KSP a precipitate of CaF2 would form. b. Example. Will a precipitate of CaF2 form when 250 ml of a 1.0x10-4M Ca(NO3)2 solution is mixed with 750 ml of a 1.0x10-3M NaF solution ? Have diluted both solutions, must get the final molarities.

[Ca2+] = 1.0x10-4( 2501000 ) = 2.5x10-5M [F-] = 1.0x10-3( 750

1000 ) = 7.5x10-4M

Q = [Ca2+][F-]2 = (2.5x10-5)(7.5x10-4)2 = 1.4x10-11 < 1.7x10-10 ∴ no precipitate will form. 2.. Under certain conditions it is possible to have a solution in which Q > KSP. Such

17

solutions are said to be supersaturated. a. A supersaturated solution is thermodynamically unstable and precipitation should occur. However, because of very slow kinetics, precipitation can not begin. These metastable solutions can sometimes last for long periods of time. b. In order for crystallization to take place the ions must come together to form micro crystallites, this process is called nucleation. Some of these micro crystallites will grow to form the macro crystals that are observable and which fall to the bottom of the vessel. If the crystal structure is complex, nucleation may not occur readily and supersaturated solutions form. c. Crystallization can be promoted by adding a small seed crystal of the solid or by scratching the sides of the contained. Scratching the container introduces small particles of glass that can serve as nucleation sites. 3. Digestion. a. When a solid precipitates there is a distribution of particle sizes. The small particles are many times difficult to filter or otherwise remove from the supernatant liquid. b. Particle size can be increased by allowing the precipitate to remain in contact with saturated solution. This process is called digestion. c. Since dissolution is a surface phenomenon, small crystals which have high surface area to mass ratios will dissolve faster than large particles. Therefore, as the solid sits in contact with the saturated solution the smaller particles will dissolve and the larger ones will grow. IV. Complex Ion Formation. A. General. 1. Complex ions are groups of atoms covalently bonded together that bear an electric

charge.

a. Some complex ions, such as, SO2-4 , NO-

3 , or OH- , are so stable that they can be

considered as single units. Others, such as Ag(NH3)+2 , are not as stable and the

dissociation process can be detected. b. Many complex ions, especially those involving a transition metal and Lewis bases, dissociate to the extent that the dissociation equilibrium constant can be measured. 2. Example.

a. The formation reaction of Ag(NH3)+2 is

Ag+ + 2NH3 Ag(NH3)+2

The formation equilibrium constant, Kf = [Ag(NH3)

+2]

[Ag+][NH3]2 = 1.5x107.

18

b. Many times the dissociation constants are listed.

Kdiss = 1

Kf = 1

1.5x107 = 6.7x10-8 = [Ag+][NH3]2

[Ag(NH3)+2]

B. Calculations involving Kf. 1. In the qualitative analysis scheme the test for Ag+ involves the fact that AgCl is insoluble

in water but dissolves in NH3, due to the formation of Ag(NH3)+2 .

Suppose that 0.01 mole of AgCl is to be dissolved in 100 ml of an NH3 solution, calculate the minimum concentration of NH3 needed to do this. There are two equilibria to work with: AgCl(s) Ag+ + Cl- KSP = [Ag+][Cl-] = 1.6x10-10

Ag+ + 2NH3 Ag(NH3)+2 Kf =

[Ag(NH3)+2]

[Ag+][NH3]2 = 1.5x107

a. If all the AgCl dissolves, there will be 0.01 mole of Cl- in the 100 ml (0.1 L) for a

[Cl-] = 0.010.10 = 0.10M.

b. The [Ag+] must be decreased to the point that 0.10M Cl- would not cause precipitation of AgCl. Therefore, the [Ag+] must be

[Ag+] = KSP[Cl-] =

1.6x10-100.10 = 1.6x10-9

c. The problem becomes: what must be the [NH3] to maintain a [Ag+] = 1.6x10-9 M,

while the [Ag(NH3)+2 ] = 0.10M ?

Kf = 1.5x107 = [Ag(NH3)

+2]

[Ag+][NH3]2 = 0.10

(1.6x10-9)[NH3]2

[NH3] = 0.10

(1.6x10-9)(1.5x107) = 2.05M

4) If one added 100 ml of 2.05M NH3 to the 0.01 mole of AgCl, it would not all dissolve. The reason is that 2.05M must be the concentration of the NH3 remaining in solution after the AgCl dissolved. The total NH3 needed is: mole of free NH3 = (2.05M)(0.10 L) = 0.205 mole mole of NH3 for complexation = (0.01)(2) = 0.02 mole total mole of NH3 required = 0.225 mole

∴ the total NH3 concentration that must be used is 0.225 mol

0.10 L = 2.25M

19

2. Calculate the molar solubility of AgCl in 6.0M NH3. a. Can combine the KSP and Kf expressions. AgCl(s) Ag+ + Cl- KSP = [Ag+][Cl-] = 1.6x10-10

Ag+ + 2NH3 Ag(NH3)+2 Kf =

[Ag(NH3)+2]

[Ag+][NH3]2 = 1.5x107

____________________________________________________________________ AgCl(s) + 2NH3 Ag(NH3)

+2 + Cl-

K = [Ag(NH3)

+2][Cl-]

[NH3]2 = ( [Ag(NH3)+2]

[Ag+][NH3]2 )([Ag+][Cl-]) = (1.6x10-10)(1.5x107) =

2.4x10-3 b. Let S = molar solubility of AgCl

[Ag(NH3)+2 ] = S, [Cl-] = S, [NH3] = 6.0-2S

[Ag(NH3)

+2][Cl-]

[NH3]2 = 2.4x10-3 = (S)(S)

(6.0-2S)2 = ( S6.0-2S )2

S

6.0-2S = 2.4x10-3 = 4.90x10-2 S = 0.27 mol / L

20

Solution Equilibrium

Acid/Base Problems

1.Fill in the missing information in the following table.

Solute Molarity [H3O+] [OH–] pH pOH HNO3 0.063 HBr 3.26 HClO4 10.51 KOH 5.0x10–9 KCl 0.087 Ba(OH)2 4.0x10–3

(ans. See last page) 2. At a temperature of 100 °C the value of Kw for water is 1.0x10 - 12.

What is the pH of pure water at 100 °C ? (ans. 6.00) 3. Calculate the pH of a 1.0x10-3 M solution of the salt NaA. HA is a weak acid

whose KA is 1x10-4. Ans: 7.5 4. Calculate the pH of 0.10M solutions of weak monoprotic acids with the

following Ka's. 1.0x10-3 b. 1.0x10-4 c. 1.0x10-5 d. 1.0x10-6 e.

1.0x10-7 f. 1.0x10-8 Answers: a. 2.0 b. 2.5 c. 3.0 d.3.5 e. 4.0 f. 4.5 5. A weak acid, HA, has a value of Ka = 1.0x10-4. Calculate the pH's of

solutions of HA that have the following molarities. a. 1.0 b. 0.10 c. 1.0x10-2 d. 1.0x10-3 e.1.0x10-4 f.

1.0x10-6 g. 1.0x10-8 Answers: a. 2.00 b. 2.50 c. 3.02 d. 3.57 e. 4.21 f. 6.00 g. 7.00 6. Calculate the pH of a 0.10 M solution of the salt [(C2H5)3NH]+Cl-.

Given (C2H5)3N is a weak base whose value of KB is 1 x 10-3. Answer 6.0

7. Oxalic acid, H2C2O4, is a diprotic acid whose K1= 5.9x10-2 and K2= 6.4x10-4. Calculate the pH, H2C2O4, HC2O4-, and C2O42- concentrations in a 2.00M H2C2O4 solution.

(pH = 0.50; [H2C2O4] = 1.68 M; [HC2O42–] = [H3O+] = 0.315 M; [C2O4

2–]

= 6.4x10–4M)

21

8. Carbonic acid, H2CO3 is a weak diprotic whose K1 = 4.2=10–7 and K2 =

4.8x10–11. Calculate the pH and the concentration of CO32– in a 0.50 M

H2CO3 solution. (Ans. 3.34, 4.8x10-11)

Buffers and Titrations

9. 50 ml of a 0.10M solution of a weak acid, HA, whose KA equals 1.0 x 10-4,

is diluted to 250 ml and then titrated with 0.10 M NaOH. Calculate pH after the following ml of the NaOH solution have been added

a) 0 b) 5 c) 25 d) 49 e) 50 f) 51 g) 60 Answers: a) 2.85 b) 3.05 c) 4.00 d) 5.70 e) 8.11 f) 10.52 g) 11.51 10. Fifty ml of 0.1M NH3 is titrated with 0.10M HCl. The KB of the weak

base NH3 is 1.8x 10-5.

Calculate the pH after the following ml of the HCl solution have been added. a) 0 b) 2 c) 10 d) 25 e) 45 f) 49 g) 50 h) 51 i) 55 Answers: a) 11.13 b) 10.61 c) 9.86 d) 9.36 e) 8.30 f) 7.56 g) 5.28

h) 3.00 i) 2.32

11. Shown below is the titration curve for the titration of 50.00 ml of a solution of the monoprotic acid, HA, with 0.01M NaOH. What is the molarity of the acid? Is the acid a weak acid or a strong one? If it is a weak acid, what is the value of Ka? ( 0.02M, weak, Ka=2.8x10-4)

2

4

6

8

10

12

pH

0 20 40 60 80 100mL NaOH

120 140 160

Titration of 50.00 mL of HA with 0.01M NaOH

22

12. Shown below is the titration curve for the titration of 50.0 ml of a solution of the weak ammonia-type base, methyl amine, CH3NH2, with 0.01M HCl. What is the molarity of the base? What is the value of Kb for methyl amine?

( 0.02 M answer given; Kb = 5.6x10-4)

2

4

6

8

10

12p

H

0 50 100 150mL 0.01M HCl

13. Consider reactions involving the following acids and bases. State in

each case whether the pH at the equivalence point would be below 7, at 7, or above 7.

a) acid = HClO4 base = KOH b) acid = HNO3 base = NH3 c) acid = HF base = LiOH 14. HNO2 is a weak acid whose Ka= 4.5x10–4. Who many grams of

NaNO2 would have to be added to 250 ml of a 0.44M HNO2 acid solution to give a buffer of pH = 4.00. (33.9 g)

15. A 36.0 ml sample of a 0.40 M HNO3 solution was titrated with 0.48 M KOH.

Calculate the pH after the following volumes of KOH have been added: a) 0 ml; b) 10.0 ml; c) 20.0 ml; 30.0 ml; e) 40.0 ml; f) 50.0 ml.

a) 0.40 b) 0.68 c) 1.07 d) 7.00 e) 12.80 f) 13.05 16. Calculate the pH’s of the following: a) the solution formed by mixing 20.0 ml of a

0.30M HCl solution with 12 ml of a 0.25 M NaOH. b) the solution formed by mixing 18.0 ml of a 0.46M HC6H7O6 solution with 10.0 ml of a 0.35M NaOH (Ka for HC6H7O6 = 8.0x10–5); c) the solution formed by mixing 15.0 ml of a 0.47M HF with 20.0 ml of a 0.20M KOH solution (Ka for HF = 7.1x10–4); d) the solution formed by mixing 30.0 ml of a 0.20 M HCHO2 solution with 22.0 ml of a 0.40 M KOH solution (Ka for HCHO2 = 1.7x10–4) e) the solution formed by dissolving

23

2.85 g of KF in 120 ml of a 0.30 M HF solution (Ka for HF = 7.1x10–4). f) the end point of the titration of 58.0 ml of a 0.48 M HCHO2 with 0.40 M KOH.

a) 1.03 b) 3.96 c) 3.03 d) 12.73 e) 4.28 f) 8.55 17. A 40.0 ml sample of a 0.40 M HNO2 solution was titrated with 0.80 M KOH. Given

that HNO2 is a weak acid whose Ka = 4.5x10–4, calculate the pH after the following volumes of KOH have been added: a) 0 ml; b) 5.0 ml; c) 10.0 ml; d) 20.0 ml;

e) 30.0 ml; f) 40.0 ml a) 1.87; b) 2.87 c) 3.35 d) 8.39 e) 13.06 f) 13.30 ) 18. Calculate the pH of the following buffers: a) 0.35 M NaIO4/0.20M

HIO4 (Ka = 5.6x10-9) b) 0.48 M NaIO/0.78 M HIO (Ka = 2.0x10-11) c) 0.40 M NH3 (Kb = 1.8x10-5)/0.55 M NH4Cl. Answers: a) 8.49 b) 10.49 c) 9.12

19.Drawn below is a plot of pH vs mL of 0.40M NaOH for the titration of

50.0 mL of a weak diprotic acid. What is the concentration of the acid? What are the values of Ka1 and Ka2 for the acid?

(Answer: 0.80M, Ka1 = 5.0x10-3, Ka2 = 6.3x10-8)

50 100 150 200 250 300 350 400

10

12

14

mL 0.4M NaOH

pH

6

4

2

0

0

8

24

KSP and Complex ions

20. The Ksp of Pb(IO3)2 at 25 °C = 3.2 x 10-13. What is the solubility of

Pb(IO3)2 in moles per liter and in grams per liter?(4.3 x 10-5 mol/L ;

g/L = 0.024) 21. Calculate the Ksp of the following compounds. The solubilities are

given in moles per liter. a) Mg(OH)2 = 1.3 x 10-4 mole/liter (8.8 x 10-12) b) Ag2C2O4 = 1.4 x 10-4 mole/liter (1.1 x 10-11) 22. Calculate the Ksp of the following salts. Solubilities are given in

grams/liter. a) BaCrO4 = 2.3 x 10--3 g/l (8.3 x 10-11) b) CaF2 = 2.7 x 10-2 g/l (1.7 x 10-10) 23. What [SO42-] must be exceeded to produce a RaSO4 precipitate in 500 ml of a

solution containing 0.00010 moles of Ra2+? (Ksp = 4 x 10-11). (2 x 10-7) 24. A solution contains an Mg2+ concentration of 0.001 mole/liter. Will Mg(OH)2 (Ksp = 8.9 x 10-12) precipitate if the OH- concentration of the solution is a) 10-5 mole/liter? b) 10-4 mole/liter? [ a) Q = 1x10–13, no b) Q = 1x10–11, yes] 25. How many grams of NaOH are required to start the precipitation of Mg(OH)2

(Ksp = 8.9 x 10-12) in 100 ml of a solution which contains 0.1 g of MgCl2? (1.16 x 10-4 g) 26. The value of Ksp for PbCl2 is 1.6 x 10-5. Will a precipitate of PbCl2

form when the following solutions are mixed? a) 100 ml of 0.01 M Pb(NO3)2 and 100 ml of 0.002 M NaCl. b) 10 ml of 0.01 M Pb(NO3)2 and 30 ml of 0.2 M NaCl. c) 10 ml of 0.01 M Pb(NO3)2 and 20 ml of 0.06 M NaCl.

d) 10 ml of 0.01 M Pb(NO3)2 and 20 ml of 0.06 M CaCl2.

[ a) no, Q = 5x10–9 b) yes, Q= 5.6x10–5 c) no, Q= 5.3x10–6 d) yes, Q=2.1x10–5] 27. Cadmium carbonate, CdCO3, is a sparingly soluble salt whose Ksp= 2.5x10-14. It is

also known that Cd2+ forms the [Cd(NH3)4]2+ complex ion that has a formation constant, Kf = 1.0x107.

a) Calculate the molar solubility of CdCO3 in 6.0M NH3. (0.018 mol/L)

25

b) What NH3 concentration would be required to dissolve 4.00 mmol of CdCO3 in one liter of the solution? (2.85M)

Answers, Question 1: Solute Molarity [H3O+] [OH–] pH pOH HNO3 0.063 6.3x10-2 1.6x10-13 1.20 12.80 HBr 5.5x10-4 5.5x10-4 1.8x10-11 3.26 10.74 HClO4 3,2x10-4 32.x10-4 3.1x10-11 3.49 10.51 KOH 2.0x10-6 5.0x10–9 2.0x10-6 5.70 8.30 KCl 0.087 1.0x10-7 1.0x10-7 7.00 7.00

Ba(OH)2 2.0x10-3 2.5x10-12 4.0x10–3 11.60 2.40