Ch. 15 & 16 Review

Everything except Polyprotics & Lewis Acids/Bases!!

Practice 1

Write the equilibrium-constant expressions for the following processes:

a) H2(g) + I2(g) 2HI(g)

b) Cd4+(aq) + 4Br-

(aq) CdBr4(aq)

c) P4 (s) + 5 O2 (g) ↔ P4O10 (s)

Practice 2

For the UNBALANCED reaction:

NH3 (aq) ↔ N2 (g) + H2 (g)

1. Write an equilibrium expression

2. Calculate the value of K at 127˚C for: [NH3] = 3.1x10-2M

[N2] = 8.5x10-1 M

[H2] = 3.1x10-3 M

3. Calculate the value of K with the above concentrations for the following reaction:

1/2 N2 + 3/2 H2 ↔ NH3

Practice 3

A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472°C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2 , 2.46 atm N2 , and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction

Practice 4

Sulfur trioxide decomposes at high temperature in a sealed container:

Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

Practice 5

At 448°C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0 10–2 mol of HI, 1.0 10–2 mol of H2, and 3.0 10–2 mol of I2 in a 2.00-L container.

Practice 6

Consider the equilibrium

In which direction will the equilibrium shift when (a)N2O4 is added(b)NO2 is removed(c)the total pressure is increased by addition of N2(g)

(d)the volume is increased(e)the temperature is decreased

7.) Determine if the following is acidic, basic, or neutral. Cu(NO3)2

Answer

Ions = Cu+2, NO3-

Cu+2 + H2O ↔ Cu(OH)2 + H+

weak base **acidic (due to

H+)

NO3- + H2O ↔ HNO3 + OH-

strong acid **neutral

(will dissociate, so H+ and OH- will form water)

Salt = Acidic

8.) Determine if the following is acidic, basic, or neutral. KClO4

Answer

Ions = K+, ClO4-

K+ + H2O ↔ KOH + H+

strong base **neutral(will dissociate, so H+ and OH- will form water)

ClO4- + H2O ↔ HClO4 + OH-

strong acid **neutral

(will dissociate, so H+ and OH- will form water) Salt = neutral

9.) Determine if the following is acidic, basic, or neutral. NaH2PO4

Answer

Ions = Na+, H2PO4-

Na+ + H2O ↔ NaOH + H+

strong base **neutral (will dissociate, so H+ and OH- will form water)

H2PO4- + H2O ↔ H3PO4 + OH-

weak acid **basic (due to

OH-)

Salt = basic

10.) Determine if the following is acidic, basic, or neutral. LiF

Answer

Ions = Li+, F-

Li+ + H2O ↔ LiOH + H+

strong base **neutral(will dissociate, so H+ and OH- will form water)

F- + H2O ↔ HF + OH-

weak acid **basic (due to

OH-)

Salt = basic

11) Determine if the following is acidic, basic, or neutral. (NH4)2CO3

Extra info: Kb (of NH3) = 1.8x10-5

ka (of HCO3-) = 5.6x10-11

Answer Ions = NH4

+, CO3-2

NH4+ + H2O ↔ NH3 + H3O+

weak base **acidic (due to H3O+)

CO3- + H2O ↔ HCO3

- + OH-

weak acid **basic (due to OH-)

**must compare Kb vs. Ka to decide pH of salt

Kb (of NH3) = 1.8x10-5

Ka of NH4+ = 5.6x10-10

ka (of HCO3-) = 5.6x10-11

kb of CO3-2 = 1.8x10-4

Salt = basic since Kb is greater than ka

12.) Complete the following table:pH [H+] pOH [OH–] Acidic, basic,

or neutral?

5.4 x 10–4

7.8 x 10-10

10.75

5.00

Answers:

ROW1

pH = 3.27

pOH = 10.73

[OH–] = 1.9 x 10–11

acidic (since pH < 7) ROW 2

pH = 4.89

[H+] = 1.3 x 10–5

pOH = 9.11

acidic (since pH < 7)

ROW 3

[H+] = 1.8 x 10-11

pOH = 3.25

[OH–] = 5.6 x 10–4

basic (since pH > 7) ROW 4

pH = 9.00

[H+] = 1.0 x 10–9

[OH–] = 1.0 x 10–5

basic (since pH > 7)

Practice 13

Calculate the pH of a 0.0430 M HNO3 solution.

Answer

Since HNO3 is a strong acid, the nitric acid solution will be 100% ionized.

Thus [H+] = [NO3–] = 0.0430 M.

The pH = - log [0.0430] = 1.37

Practice 14

Calculate the pH of a 0.020 M Ba(OH)2 (aq) solution.

Answer

Since Ba(OH)2 is a strong base it is 100% ionized. Note that ionization gives 2 OH– ions for each mole of Ba(OH)2.

Thus [OH–] = 2 x 0.020 M = 0.040 M pOH = -log[0.040] = 1.40

pH = 14 – 1.40 = 12.60

Practice 15

Calculate the pH of a 0.250 M HC2H3O2 solution. Ka(HC2H3O2) = 1.8 x 10-5.

Answer

Ka [H][C2H3O2

]

[HC2H3O2 ]1.8 10 5

x 2

0.250 - x

x2

0.250

Answer:

Balanced Equation HC2H3O2 H+ + C2H3O2–

Initial Concentration (M) 0.250 0 0

Change (M) - x x x

Equilibrium Concentration (M) 0.250 - x x x

(a)

Thus x2 = 4.5 x 10-6; x = 2.12 x 10-3 = [H+]. pH = 2.67.

Practice 16

Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb = 4.4 x 10–4.

AnswerAnswer:

Since CH3NH2 is a weak base, the balanced equation for the reaction is CH3NH2 + H2O CH3NH3+ + OH–.

Balanced Equation CH3NH2

CH3NH3+ + OH–

 

Initial Concentration (M) 0.600

___0 0

Change (M) - x

___x x

Equilibrium Concentration (M) 0.600 - x

___x x

Kb [BH][OH ]

[B]

[CH3NH3][OH ]

[CH 3NH 2]

x2

0.600 x

x2

0.6004.4 10 4

Thus x = 1.62 x 10-2 = [OH–], and pOH = 1.79. The pH = 12.21.

Practice 17

The pH of a 0.10 M solution of a weak base is 9.67. What is the Kb of the base?

Answer

[OH ] 10 pOH 104.33 4.68 10 5 x

82522

102.210.0

)1068.4(

10.010.0[B]

]][OH[BH

x

x

xKb

Answer:The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is given below.

Balanced Equation B

BH+ + OH–

 

Initial Concentration (M) 0.10

___0 0

Change (M) - x

___x x

Equilibrium Concentration (M) 0.10 - x

___x x

At equilibrium, [OH–] = [BH+] = x. Use the pH to calculate the [OH–] at equilibrium (which is the value of x). Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus :

Practice 18

Use the following acidity constants to help answer the questions below:

Ka(HC2H3O2) = 1.8 x 10– 5

Ka(HCN) = 4.9 x 10–10 Ka(HCOOH) = 1.7 x 10-4

1. Which of the three acids is the weakest?

2. Which of the following bases is the strongest: C2H3O2

-, CN- , or HCOO- ?

3. What is the pKa of HCN?

4. What is the Kb for CN- ?

Answer

(1) smallest k value = HCN (2) strong base = weakest acid = lowest k

value = HCN (acid) = CN- (base)

(3) pKa = -log Ka = -log (4.9x10-10) = 9.31 (4) Ka x Kb = 1x10-14

kb = 1x10-14 = 2.0 x 10-5

4.9x10-10

19.) Predict whether an aqueous solution would be acidic, basic or neutral?

1.sodium nitrate

2.ammonium iodide

3.sodium bicarbonate

4.ammonium cyanide

5.sodium hypochlorite

6.potassium acetate

Predict whether an aqueous solution would be acidic, basic or neutral? ANSWERS

1.sodium nitrate = NaNO3 = neutral

2.ammonium iodide = NH4I = acidic

3.sodium bicarbonate = NaHCO3 = basic

4.ammonium cyanide = NH4CN = basic (Kb ↑)

kb (for NH3) vs. Ka (for HCN) 1.8x10-5 vs. 4.9x10-10

5.sodium hypochlorite = NaClO = basic

6.potassium acetate = KC2H3O2 = basic

Practice 20 (from ch. 15 study questions)

Suppose that 0.50 moles of hydrogen gas, 0.50 moles of iodine gas, and 0.75 moles of hydrogen iodide gas are introduced into a 2.0 Liter vessel and the system is allowed to reach equilibrium.

H2(g) + I2(g) ↔ 2 HI(g)

Calculate the concentrations of all three substances at equilibrium. At the temperature of the experiment, Kc equals 2.0 x 10-2.

Practice 21 (from ch. 15 study questions) Nitrosyl chloride NOCl decomposes to nitric oxide and chlorine

when heated:

2 NOCl(g) ↔ 2 NO(g) + Cl2(g)

At 600K, the equilibrium constant Kp is 0.060. In a vessel at 600K, there is a mixture of all three gases. The partial pressure of NOCl is 675 torr, the partial pressure of NO is 43 torr and the partial pressure of chlorine is 23 torr.

a. What is the value of the reaction quotient?

b. Is the mixture at equilibrium?

c. In which direction will the system move to reach equilibrium?

d. When the system reaches equilibrium, what will be the partial pressures of the components in the system? (just set up the problem, do not solve all the way!!)