Chapter 16 16.1 HBr is a strong acid and will …Chapter 16 506 16.1 HBr is a strong acid and will therefore give its hydrogen atom to a water molecule, making a hydronium cation and

Chapter 16

506

16.1 HBr is a strong acid and will therefore give its hydrogen atom to a water molecule, making a hydronium cation and a bromine anion.

16.2 HNO3 is a strong acid and will therefore give its hydrogen atom to a water molecule,

making a hydronium cation and a nitrate anion.

16.3 HClO4 is a strong acid and dissolves in water to generate H3O+cations and ClO4

− anions. Since there are no bases present for the hydronium ion to react with, the only equilibrium occurring is proton transfer reaction between water molecules:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14 To determine the concentrations of hydroxide and hydronium ions, set up a concentration table, write the equilibrium expression, and solve for the final concentrations.

The initial concentration of hydronium ions is the same as the concentration of perchloric acid, [H3O+] = 1.25 x 10-3M. The concentration table is:

Reaction: H2O + H2O H3O+ + OH–

Start (M) ---- ---- 1.25 x 10-3 0

Change (M) ---- ---- + x + x

Final (M) solvent solvent 1.25 x 10-3 + x x

The equilibrium expression is: Kw = [H3O+][OH-]

1.00 x 10-14 = [0.00125 + x][x], We can make the assumption that x << 0.00125M 1.00 x 10-14 = [0.00125][x], x = 8.00 x 10-12 M = [OH-]; [H3O+] = 1.25 x 10-3 M + 8.00 x 10-12 M = 1.25 x 10-3 M Thus, the assumption that x<<0.00125 is valid.

Chapter 16

507

16.4 NaOH is a strong base and dissolves in water to generate Na+ cations and OH- anions. Since there are no acids present to react with the hydroxide, the only equilibrium occurring is proton transfer reaction between water molecules:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14 To determine the concentrations of hydroxide and hydronium ions set up a concentration table, write the equilibrium expression, and solve for the final concentrations. The initial concentration of hydroxide ions will be the concentration of NaOH, [OH-] = 3.65 x 10-2M. The concentration table is:


Start (M) ---- ---- 0 0.0365

Change (M) ---- ---- + x + x

Final (M) solvent solvent x 0.0365 + x

The equilibrium expression is:

Kw = [H3O+][OH-] 1.00 x 10-14 = [x][0.0365 + x]

We can make the assumption that x << 0.0365 1.00 x 10-14 = [x][0.0365], x = 2.74 x 10-13 M = [H3O+]; our assumption is valid

[OH-] = 0.0365M + 2.74 x 10-13 M = 0.0365 M 16.5 We are asked to determine the final concentrations of all the ions in a final solution.

Begin by analyzing the chemistry. HCl is strong acid and dissolves in water to generate H3O+cations and Cl- anions. Any water solution always has OH- and H3O+ ions with the equilibrium:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14 Therefore, the ions present in this solution are: H3O+, OH-, and Cl-

This is a dilution type problem. The first step is to determine the concentration of HCl in the flask after the dilution.

MiVi = MfVf

Mf = mL 100.

mL) M)(1.00 (12.1

f

ii =V

VM = 0.121 M HCl in final solution.

Since Cl- is a spectator ion, its concentration is the same as that of HCl, [Cl-] = 0.121 M.

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508

The rest of the ion concentrations are determined by the equilibrium. Set up a concentration table, write the equilibrium expression, and solve for the final concentrations.


Start (M) ---- ---- 0.121 0

Change (M) ---- ---- + x + x

Final (M) solvent solvent 0.121 + x x


Kw = [H3O+][OH-] 1.00 x 10-14 = x(0.121 +x ); assume x << 0.121

1.00 x 10-14 = x(0.121), x = [OH-] = 8.26 x 10-14M

[H3O+] = 0.121 M + 8.26 x 10-14M = 0.121 M Here are the final concentrations: [Cl-] = [H3O+] = 0.121 M

[OH-] = 8.26 x 10-14 M 16.6 We are asked to determine the final concentrations of all the ions in a final solution.

Begin by analyzing the chemistry. HClO4 is a strong acid and dissolves in water to generate H3O+cations and ClO4

− anions. Any water solution always has OH- and H3O+ ions with the equilibrium:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14 Therefore, the ions present in this solution are: H3O+, OH-, and ClO4

- This is a dilution type problem. The first step is to determine the concentration of HClO4 in the flask after the dilution.

MiVi = MfVf

Mf = mL 1000.

mL) M)(5.00 (14.8

f

ii =V

VM = 0.0740 M HClO4 in final solution.

Since ClO4− is a spectator ion, its concentration is the same as that of HClO4,

[ ClO4− ]= 0.0740M.

The rest of the ion concentrations are determined by the equilibrium. Set up a concentration table, write the equilibrium expression, and solve for the final concentrations.

Chapter 16

509


Start (M) ---- ---- 0.0740 0

Change (M) ---- ---- +x +x



Kw = [H3O+][OH-] 1.00 x 10-14 = x(0.0740 +x ); assume x<<0.0740

1.00 x 10-14 = x(0.0740), x = [OH-] = 1.35 x 10-13 M [H3O+] = 0.0740 M + 1.35 x 10-13 M = 0.0740 M

Here are the final concentrations: [ClO4

-] = [H3O+] = 0.0740 M [OH-] = 1.35 x 10-13 M

16.7 We are asked to determine the concentrations of hydroxide and hydronium ions in the

solution. Begin by analyzing the chemistry. HCl is a strong acid and dissolves in water to generate H3O+cations and Cl- anions. Any water solution always has OH- and H3O+ ions with the equilibrium:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14

The first step is to determine the initial number of moles of HCl gas and convert that to concentration of HCl dissolved in the solution:

MMHCl = 1.008 g/mol + 35.453 g/mol = 36.461 g/mol

0.488 g

g 36.461mol 1 =0.01338 mol HCl dissolved.

[HCl] = L 0.325mol 0.01338 = 0.0412 M

To determine the concentrations of the ions, construct a concentration table, write the equilibrium expression, and solve for the final concentrations. The initial concentration of hydronium ions will be the same as the concentration of HCl, [H3O+] = 0.0412 M.

Chapter 16

510


Start (M) ---- ---- 0.0412 0

Change (M) ---- ---- +x +x


The equilibrium expression is: Kw = [H3O+][OH-] 1.00 x 10-14 = (0.0412 +x)x; Assume x << 0.0412 1.00 x 10-14 = (0.0412)x; x = 2.43 x 10-13 M = [OH-]; the assumption is vaild [H3O+] = 0.0412 M + 2.43 x 10-13 M = 0.0412 M

16.8 We are asked to determine the concentrations of hydroxide and hydronium ions in the

solution. Begin by analyzing the chemistry. NaOH is a strong base and dissolves in water to generate Na+ cations and OH- anions. Any water solution always has OH- and H3O+ ions with the equilibrium:

H2O + H2O OH- + H3O+ Kw = 1.00 x 10-14

The first step is to determine the concentration of NaOH: MMNaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol

0.345 g

g 40.00mol 1 = 0.008625 mol NaOH dissolved.

[NaOH] = L 0.225mol 0.008625 = 0.03833 M


Start (M) ---- ---- 0 0.03833

Change (M) ---- ---- +x +x

Final (M) Solvent solvent x 0.03833 + x

Kw = [H3O+][OH-]; 1.00 x 10-14 = x(0.03833 +x ) assume x << 0.03833

Chapter 16

511

1.00 x 10-14 = x(0.03833); x = 2.61 x 10-13 M = [H3O+]; assumption is valid [OH-] = 0.03833 M + 2.61 x 10-13 M = 0.03833 M

16.9 Conversion from hydronium ion molarity to pH is accomplished by taking logarithm to

base ten and changing sign, pH = – (log[H3O+]): (a) – 0.60; (b) 5.426; (c) 2.32; and (d) 3.593.

16.10 Conversion from hydronium ion molarity to pH is accomplished by taking logarithm to

base ten and changing sign, pH = – (log[H3O+]): (a) -0.097; (b) 5.530; (c) 1.437; and (d) 3.128 .

16.11 Because pH + pOH = 14.00, pH = 14.00 – pOH. Convert from hydroxide ion molarity to

pOH by taking logarithm to base ten and changing sign, pOH = – (log[OH-]). Thus, pH = 14 + log[OH-]:

(a) 14.60 ; (b) 8.574; (c) 11.68; and (d) 10.407. 16.12 Because pH + pOH = 14.00, pH = 14.00 – pOH. Convert from hydroxide ion molarity to

pOH by taking logarithm to base ten and changing sign, pOH = – (log[OH-]). Thus, pH = 14 + log[OH-]:

(a) 14.097 ; (b) 8.470; (c) 12.563; and (d) 10.872. 16.13 Take 10-pH to convert pH into hydronium ion concentration:

(a) 0.22 M; (b) 1.4 x 10-8 M; (c) 2.1 x 10-4 M; (d) 4.7 x 10-15 M. 16.14 Take 10-pH to convert pH into hydronium ion concentration: (a) 0.028 M; (b) 1.4 x 10-4 M; (c) 1.8 x 10-10 M; (d) 6.0 x 10-12 M. 16.15 Use pH + pOH = 14.00 to convert pH to pOH. Then take 10-pOH to convert pOH into

hydroxide ion concentration: (a) pOH = 13.34, [OH-] = 4.6 x 10-14 M;

(b) pOH = 6.15, [OH-] = 7.1 x 10-7 M; (c) pOH = 10.32, [OH-] = 4.8 x 10-11 M;

(d) pOH = – 0.33, [OH-] =2.1 M. 16.16 Use pH + pOH = 14.00 to convert pH to pOH. Then take 10-pOH to convert pOH into

hydroxide ion concentration: (a) pOH = 12.44, [OH-] = 3.6 x 10-13 M;

(b) pOH = 10.15, [OH-] = 7.1 x 10-11 M; (c) pOH = 4.25, [OH-] = 5.6 x 10-5 M;

(d) pOH = 2.78, [OH-] = 1.7 x 10-3 M. 16.17 To calculate the pH of a solution, it is necessary to determine either the hydronium ion

concentration or the hydroxide ion concentration.

Chapter 16

512

For each species determine the initial concentrations, construct a concentration table, write the equilibrium expression, and solve for the concentrations.

(a) Strong base, carry out the water equilibrium to determine H3O+

Reaction: H2O + H2O OH- + H3O+

Start (M) ---- ---- 1.5 0

Change (M) ---- ---- + x + x


The equilbrium expression is: Kw = 1.0 x 10-14 = [H3O+][OH-] = x(1.5 + x); assume x << 1.5 1.0 x 10-14 = x(1.5); from which x = 6.7 x 10-15 M = [H3O+]; assumption valid pH = -log(6.7 x 10-15) = 14.18.

(b) weak base, carry out equilibrium calculation to determine [OH-]:

Reaction: H2O + C5H5N C5H5NH+ + OH–

Start (M) ---- 1.5 0 0

Change (M) ---- -x +x +x

Final (M) Solvent 1.5 – x x x


Kb = 1.7 x 10-9 = x

x−

=5.1N]H[C

][OH]NHH[C 2

eq55

eq-

eq+

55 ; Assume x << 1.5:

1.7 x 10-9 = 1.5

2x, from which x2 = 2.55 x 10-9 and x = 5.0 x 10-5; assumption is valid

[OH-] = 5.0 x 10-5 M, pOH = -log (5.0 x10-5) = 4.30, and pH = 14.00 – 4.30 = 9.70;

Chapter 16

513

(c) weak base, carry out equilibrium calculation to determine [OH-]:

Reaction: H2O + NH2OH NH3OH + + OH–

Start (M) ---- 1.5 0 0


Final (M) solvent 1.5 – x x x


Kb = 8.7 x 10-9 = x

x−

=5.1OH][NH

][OH]OH[NH 2

eq2

eq-

eq+

3 ; Assume x << 1.5:

8.7 x 10-9 = 1.5

2x, from which x2 = 1.31 x 10-8 and x = 1.1 x 10-4; assumption valid

[OH-] = 1.1 x 10-4 M, pOH = -log(1.1 x 10-4) = 3.96, and pH = 14.00 – 3.96 = 10.04;

(d) Weak acid, carry out equilibrium calculation to determine [H3O+]:

Reaction: H2O + HCO2H HCO2- + H3O+

Start (M) ---- 1.5 0 0

Change (M) ---- - x + x + x

Final (M) solvent 1.5 - x x x


Ka= 1.8 x 10-4 = x

x−

=5.1H][HCO

]O[H][HCO 2

eq2

eq+

3eq-

2 ; Assume x << 1.5:

1.8 x 10-4 = 1.5

2x, from which x2 = 2.7 x 10-4 and x = 1.6 x 10-2; assumption is valid

[H3O+] = 1.6 x 10-2 M, and pH = -log(1.6 x 10-2) = 1.80

16.18 To calculate the pH of a solution, it is necessary to determine either the hydronium ion

concentration or the hydroxide ion concentration.

Chapter 16

514

For each species determine the initial concentrations, construct a concentration table, write the equilibrium expression, and solve for the concentrations.

(a) weak base, carry out equilibrium calculation to determine [OH-]:

Reaction: H2O + NH3 OH - + NH4+

Start (M) ---- 2.5 x 10-2 0 0

Change (M) ---- -x + x + x

Final (M) solvent 2.5 x 10-2- x x x


Kb = 1.8 x 10-5 = x

x−

= 2-

2

eq3

eq-

eq+

4

10 x 5.2][NH][OH][NH

; Assume that x << 2.5 x 10-2

1.8 x 10-5 =2-

2

10 x 5.2x

x = 6.7 x 10-4 M [OH-] = 6.7 x 10-4 M, pOH = -log(6.7 x 10-4) = 3.17, and pH =10.83;

(b) weak acid, carry out an equilibrium calculation to determine [H3O+]:

Reaction: H2O + HClO ClO - + H3O+

Start (M) ---- 2.5 x 10-2 0 0




Ka= 4.0 x 10-8 = x

x−

= 2-

2

eq

eq+

3eq-

10 x 5.2[HClO]]O[H][ClO

; Assume x << 2.5 x 10-2:

4.0 x 10-8 = )10 x (2.5 2-

2x , from which

x2 = 1.0 x 10-9 and x = 3.2 x 10-5; assumption is valid

[H3O+] = 3.2 x 10-5 M, and pH = -log(3.2 x 10-5) = 4.49;

(c) weak acid, carry out an equilibrium calculation to determine [H3O+]:

Chapter 16

515

Reaction: H2O + HCN CN- + H3O+

Start (M) ---- 2.5 x 10-2 0 0




Ka= 6.2 x 10-10 = x

x−

= 2-

2

eq

eq+

3eq-

10 x 5.2[HCN]]O[H][CN

; Assume x << 2.5 x 10-2:

6.2 x 10-10 = 2-

2

10 x 2.5x , from which

x2 = 1.55 x 10-11 and x = 3.9 x 10-6; assumption is valid

[H3O+] = 3.9 x 10-6 M, and pH = -log(3.9 x 10-6) = 5.41;

(d) Strong base. The initial hydroxide concentration will be twice the concentration of Ba(OH)2 since there are two hydroxide ions per molecule. [OH-] = 2(2.5 x 10-2 M) = 5.0 x 10-2 M,

Reaction: H2O + H2O OH- + H3O+

Start (M) ---- ---- 5.0 x 10-2 0

Change (M) ---- ---- + x + x

Final (M) solvent Solvent 5.0 x 10-2 + x x

The equilbrium expression is: Kw = 1.0 x 10-14 = [H3O+][OH-] = x(5.0 x 10-2 + x); assume x<<5.0 x 10-2 1.0 x 10-14 = x(5.0 x 10-2); from which x = 2.0 x 10-13 M = [H3O+]; assumption valid pH = -log(2.0 x 10-13) = 12.70.

16.19 HONH2 is a weak base and will therefore take a proton from water to form hydroxide

ions.

Chapter 16

516

HCO2H is a weak acid and will give a proton to water to form hydronium ions.

16.20 NH3 is a weak base and will take a proton from water to form hydroxide ions:

HCN is a weak acid and will give a proton to water to form hydronium ions.

16.21 Follow standard procedures for dealing with equilibrium problems: (a) HN3 is a weak acid. Major species: HN3 and H2O, Minor species: N3- and H3O+, and

OH-

(b) Construct the concentration table, write the equilibrium expression, and solve for the concentrations.

Reaction: H2O + HN3 N3- + H3O+

Start (M) ---- 1.50 0 0

Change (M) ---- - x + x + x


Ka= 2.5 x 10-5 = eq3

eq+

3eq-3

][HN]O[H][N

= - x.

x501

2

; assume that x << 1.50

2.5 x 10-5 = 501

2

.x ;

x2 =3.75 x 10-5 and x = 6.1 x 10-3 [H3O+] = [N3-] = 6.1 x 10-3 M, and [HN3] = 1.50 M – 6.1 x 10-3 M = 1.50 M

[OH-] = 3-

-14

10 x 1.610 x 0.1

= 1.6 x 10-12 M

(c) pH = – log (6.1 x 10-3) = 2.21;

Chapter 16

517

(d) The dominant equilibrium is proton transfer from water to HN3:

16.22 Follow standard procedures for dealing with equilibrium problems: (a) HCNO is a weak acid. Major species: HCNO and H2O, Minor species: CNO- and

H3O+ and OH-; (b) Construct the concentration table, write the equilibrium expression, and solve for the

concentrations.

Reaction: H2O + HCNO CNO - + H3O+

Start (M) ---- 0.0275 0 0

Change (M) ---- - x + x + x


pKa = 3.46, Ka= 3.5 x 10-4 = eq

eq+

3eq-

[HCNO]]O[H][CNO

; Substitute and solve for x:

3.5 x 10-4 = x

x−0275.0

2, so x2 = (3.5 x 10-4)(0.0275 – x) = 9.6 x 10-6 – 3.5 x 10-4 x;

0 = x2 + 3.5 x 10-4 x – 9.6 x 10-6;

x = 362442

10 x 9.22

)10 x 4(9.6)10 x (3.5)10 x (3.52

4 −−−−

=−±−=−±−a

acbb ;

[H3O+] = [CNO-]= 2.9 x 10-3 M, and [HCNO] = 0.0275 – 0.0029 = 0.0246 M

[OH-] = 3-

-14

10 x .9210 x 1.00 = 3.4 x 10-12 M

(c) pH = – log( 2.9 x 10-3) = 2.54; (d)

16.23 Follow standard procedures for dealing with equilibrium problems:

Chapter 16

518

(a) N(CH3)3 is a weak base. Major species: N(CH3)3, H2O, Minor species: HN(CH3)3+,OH-, and H3O+;


Reaction: H2O + N(CH3)3 OH - + HN(CH3)3+

Start (M) ---- 0.350 0 0

Change (M) ---- - x + x + x



Kb = 6.5 x 10-5 = x

x−

=+

350.0])[N(CH][OH])[HN(CH 2

eq33

eq-

eq33 ; assume x << 0.350

6.5 x 10-5 = 350.0

2x ;

x2 = 2.28 x 10-5 and x = 4.8 x 10-3; assumption is valid [OH-] = [HN(CH3)3+] = 4.8 x 10-3 M,

[N(CH3)3] = 0.350 – 4.8 x 10-3 = 0.345 M;

[H3O+] = 3-

-14

10 x .8410 x 1.0 = 2.1 x 10-12 M

(c) pH =-log(2.1 x 10-12) = 11.68; (d) The dominant equilibrium is proton transfer from water to trimethylamine:

16.24 Follow standard procedures for dealing with equilibrium problems: (a) C6H5NH2 is a weak base. Major species: C6H5NH2, H2O, Minor species:

C6H5NH+,OH- , and H3O+;


Chapter 16

519

Reaction: H2O + C6H5NH2 OH - + C6H5NH3+

Start (M) ---- 1.85x10-3 0 0

Change (M) ---- -x + x + x

Final (M) solvent 1.85x10-3– x x x

Kb = 7.4 x 10-10 =x

x−

=+−

3-

2

256

356

10 x 85.1]NHH[C]NHH][C[OH ; assume x << 1.85 x 10-3

7.4 x 10-10 = 3-

2

10 x 85.1x , from which x2 = 1.37 x 10-12;

x = 1.2 x 10-6; assumption correct

[OH-] = [C6H5NH3+] = 1.2 x 10-6 M, [C6H5NH2] = 1.85 x 10-3– 1.2 x 10-6 = 1.85 x 10-3 M;

[H3O+] = 6-

-14

10 x .2110 x 1.00 = 8.3 x 10-9 M

(c) pH =-log(8.3 x 10-9) = 8.08;

(d)

16.25 Examine the chemical formula to determine the nature of a compound: (a) weak base;

(b) weak acid, (c) weak acid; (d) strong base. 16.26 Examine the chemical formula to determine the nature of a compound: (a) strong acid; (b) weak acid; (c) amphiprotic, weak acid and weak base; (d) weak acid; (e) weak base. 16.27 The conjugate base will have one less H and one less charge, a conjugate acid will have

one more H and one higher charge: Problem 16.17

C5H5N conjugate acid is C5H5NH+; HONH2 conjugate acid is HONH3

+; HCO2H conjugate base is HCO2

-;

Chapter 16

520

16.28 The conjugate base will have one less H and one less charge, a conjugate acid will have one more H and one higher charge:

Problem 16.18 Problem 16.26 NH3 conjugate acid is NH4

+; CH3CO2H conjugate base is CH3CO2-;

HClO conjugate base is ClO-; HOH conjugate base is OH-, HCN conjugate base is CN-; HOH conjugate acid is H3O+; HOCl conjugate base is OCl-;

NH2OH conjugate acid is NH3OH+. 16.29 Conjugate pairs are connected. Any aqueous solution always has OH- and H3O+ ions

with the equilibrium:

(a) NH3 is a weak base:

(b) HCNO is a weak acid:

(c) HClO is a weak acid:

(d) Ba(OH)2 is a strong base:

Chapter 16

521

16.30 Any aqueous solution always has OH- and H3O+ ions with the equilibrium:

(a) HNO3 is a strong acid:

(b) CH3CO2H is a weak acid:

(c) HOH is water:

(d) HOCl is a weak acid:

(e) NH2OH is a weak base:

16.31 Follow standard procedures for dealing with equilibrium problems: (a) The compound is a salt, so major species are Na+, SO32-, and H2O; (b) The species with acid-base properties are SO32- (a weak base) and H2O, so the

dominant equilibrium is: H2O(l) + SO32-(aq) HSO3-(aq) + OH-(aq)

(c) Construct the concentration table, write the equilibrium expression, and carry out an equilibrium calculation to determine [OH-]:

Chapter 16

522

Reaction: H2O + SO32- HSO3- + OH-

Start (M) --- 0.45 0 0

Change (M) --- – x + x + x


The equilibrium reaction is a weak base proton transfer. HSO3- is the species resulting

from the gain of a proton from SO32-, so use Ka2 to determine pKb = 14.00 – pKa:

pKa = 7.20, pKb= 6.80; The equilibrium expression is:

Kb = 1.6 x 10-7 = x

x−

=45.0][SO

][OH][HSO 2

eq-2

3

eq-

eq-3 ; Assume x << 0.45:

1.6 x 10-7 = 0.45

2x , from which x2 = 7.2 x 10-8 and x = 2.7 x 10-4; assumption valid

[OH-] = 2.7 x 10-4 M, pOH = -log(2.7 x 10-4) = 3.57, and pH = 14.00 – 3.57 = 10.43.

16.32 Follow standard procedures for dealing with equilibrium problems: (a) The compound is a salt, so major species are Na+, C6H5CO2

-, and H2O; (b) The species with acid-base properties are C6H5CO2- (a weak base) and H2O, so the

dominant equilibrium is: H2O(aq) + C6H5CO2- (aq) C6H5CO2H(aq) + OH-(aq)

(c) Construct the concentration table, write the equilibrium expression, and carry out an equilibrium calculation to determine [OH-]:

Reaction: H2O + C6H5CO2- C6H5CO2H + OH-

Start (M) --- 6.75 x 10-3 0 0

Change (M) --- – x + x + x

Final (M) solvent 6.75 x 10-3 – x x x

The equilibrium reaction is a weak base proton transfer. C6H5CO2H is the species

resulting from the gain of a proton from C6H5CO2-, so use Ka to determine

pKb = 14.00 – pKa: pKa = 4.20, pKb= 9.80;

Chapter 16

523


Kb = 1.6 x 10-10 = x

x−

=00675.0]COH[C

][OHH]COH[C 2

eq-256

eq-

eq256 ; Assume x << 0.00675:

1.6 x 10-10 = 00675.0

2x, from which

x2 = 1.1 x 10-12 and x = 1.0 x 10-6; assumption valid

[OH-] = 1.0 x 10-6 M, pOH = -log(1.0 x 10-6) = 6.00, and pH = 14.00 - 6.00 = 8.00.

16.33 Follow standard procedures for dealing with equilibrium problems: (a) The compound is a salt, so major species are NH4+, NO3-, and H2O; (b) The species with acid-base properties are NH4+ (a weak acid) and H2O, so the

dominant equilibrium is: H2O(l) + NH4+(aq) NH3(aq) + H3O+(aq) (c) carry out equilibrium calculation to determine [H3O+]:

Reaction: H2O + NH4+ NH3 + H3O+

Start (M) --- 0.0100 0 0

Change (M) --- – x + x + x


pKa = 9.25, Ka= 5.6 x 10-10 = x

x−

=0100.0][NH

]O[H][NH 2

eq3

eq+

3eq+4 ; Assume x << 0.0100:

5.6 x 10-10 = 0.0100

2x , from which

x2 = 5.6 x 10-12 and x = 2.37 x 10-6; assumption valid [H3O+] = 2.37 x 10-6 M, and

pH = -log(2.37 x 10-6) = 5.63. 16.34 Follow standard procedures for dealing with equilibrium problems: (a) The compound is a salt, so major species are NH4

+ , Br-, and H2O;

(b) The species with acid-base properties are NH4+ (a weak acid) and H2O, so the

dominant equilibrium is: H2O(l) + NH4+(aq) NH3(aq) + H3O+(aq) (c) carry out an equilibrium calculation to determine [H3O

+]:

Chapter 16

524


Start (M) --- 4.75 x 10-2 0 0

Change (M) --- – x + x + x

Final (M) solvent 4.75 x 10-2 – x x x

The equilibrium reaction is a weak acid proton transfer, NH3 is the species resulting from

the loss of a proton from NH4+ , so use Kb to determine pKa =14.00 - pKb:

pKb = 4.75, pKa= 9.25,

Ka = 5.6 x 10-10 =

[NH3]eq[H3O+]eq

[NH4+]eq -x.

x04750

2

= ; Assume x << 0.0475:

5.6 x 10-10 =0475.0

2x , from which x2 = 2.7 x 10-11 and x = 5.2 x 10-6; assumption valid

[H3O+] = 5.2 x 10-6 M,

pH = -log(5.2 x 10-6) = 5.28. 16.35 H2O + SO2 H2SO3

H2SO3 + H2O HSO3- + H3O+ (determines pH)

HSO3- + H2O SO3

2- + H3O+

16.36 H2O + CO2 H2CO3

H2CO3 + H2O HCO3- + H3O+ (determines pH)

HCO3- + H2O CO3

2- + H3O+

16.37 FeO(s) + 2 H3O+ (aq) Fe2+ (aq) + 3 H2O(l) 16.38 Fe2O3 (s) + 6 H3O + (aq) 2 Fe3+(aq) + 9 H2O(l) 16.39 (a) H2SO4 is stronger because anions are poorer proton donors than neutral species.

(b) HClO is stronger because Cl is a more electronegative atom than I. A higher electronegativity means that Cl attracts more of the electron density around it than I, weakening the H-X bond and making it easier to break (hence a better proton donor/acid).

Chapter 16

525

(c) HClO2 is stronger. O atoms are highly electronegative and attract electron density around them. Having two O atoms, HClO2 will have less electron density in the H-X bond than HClO, thus, making the bond easier to break.

16.40 (a) HBrO3 is stronger. O atoms are highly electronegative and attract electron

density around them. Having three O atoms, HBrO3 will have less electron density in the H-X bond than HBrO2, thus making the bond easier to break and making it a stronger acid.

(b) H2S is stronger because the larger S atom provides less bond overlap with the H making it easier to remove the H.

(c) H2S is a stronger acid because the negative charge on HS- makes it harder to remove the proton.

16.41 Use arrows of different sizes to show differences in electron density shifts.

H

O Cl

H

O I(b)

H

O Cl

H

O Cl

O

(c)

16.42 Use arrows of different sizes to show differences in electron density shifts.

H

O H(b)

H

S H

16.43 This problem asks for the concentration of all ionic species present in the solution. Begin by analyzing the chemistry. NaC2H3O2 is a salt that dissolves in solution to form Na+ and C2H3O2

-. Acetate is a weak base with equilibrium: H2O + C2H3O2

- C2H3O2H + OH- Kb= 5.6 x 10-10 Every aqueous solution has the water equilibrium: H2O + H2O H3O+ + OH- Kw = 1.0 x 10-14

Thus, the ionic species present in the solution are: Na+, C2H3O2

-, H3O+, and OH-

Chapter 16

526

Na+ is a spectator ion and will have the same concentration as the initial salt, [Na+] = 0.250 M

For the remaining ions we will need to set up concentration tables, write the equilibrium expressions, and solve for the ionic concentrations.

Reaction: H2O + C2H3O2

- C2H3O2H + OH-

Start (M) ---- 0.250 0 0

Change (M) ---- -x + x + x

Final (M) ---- 0.250-x x x

The acid-base equilibrium expression is:

Kb = 5.6 x 10-10 = x

x−250.0

2

; assume that x << 0.250

5.6 x 10-10 = 250.0

2x; for which x2 = 1.4 x 10-10 and x = 1.2 x 10-5; assumption valid

[C2H3O2H] = [OH-] = 1.2 x 10-5 M [C2H3O2

-] = 0.250 M – 1.2 x 10-5 M = 0.250 M Now use a second concentration table to determine the concentrations of hydronium and hydroxide ions:

Reaction: H2O + H2O H3O+ + OH-

Start (M) ---- ---- 0 1.2 x 10-5

Change (M) ---- ---- + x + x

Final (M) solvent solvent x 1.2 x 10-5 + x

Kw = 1.00 x 10-14 = (x)(1.2 x 10-5 + x); assume x << 1.2 x 10-5 1.00 x 10-14 = (x)(1.2 x 10-5); x = 8.3 x 10-10 M = [H3O+]; assumption valid [OH-] = 1.2 x 10-5 M The ionic concentrations are: [Na+] = [C2H3O2

-] = 0.250 M [OH-] = 1.2 x 10-5 M [H3O+] = 8.3 x 10-10 M

16.44 This problem asks for the concentration of all ionic species present in the solution. Begin by analyzing the chemistry. KBrO is a salt that dissolves in solution to form K+ and BrO-. BrO- is a weak base with equilibrium:

H2O + BrO- HBrO + OH- Kb= 3.6 x 10-6

Chapter 16

527

Every aqueous solution has the water equilibrium: H2O + H2O H3O+ + OH- Kw = 1.00 x 10-14

Thus, the ionic species present in the solution are: K+, BrO-, H3O+, and OH- K+ is a spectator ion and will have the same concentration as the initial salt,

[K+] = 3.45 x 10-2 M

For the remaining ions we will need to set up concentration tables, write the equilibrium expressions, and solve for the ionic concentrations.

Reaction: H2O + BrO- HBrO + OH-

Start(M) ---- 3.45 x 10-2 0 0

Change (M) ---- -x + x + x

Final (M) ---- 3.45 x 10-2 -x x x


Kb = 3.5 x 10-6 = x

x−2-

2

10 x 45.3; assume that x << 3.5 x 10-2

3.5 x 10-6 = 2-

2

10 x 45.3x

;

x = 3.5 x 10-4 M = [HBrO] = [OH-] [BrO-] = 3.45 x 10-2 M – 3.5 x 10-4 M = 3.42 x 10-2 M Now use a second concentration table to determine the concentrations of hydronium and hydroxide ions:


Start(M) ---- ---- 0 3.5 x 10-4

Change (M) ---- ---- + x + x

Final (M) solvent solvent x 3.5 x 10-4 + x

Kw = 1.00 x 10-14 = (x)(3.5 x 10-4 + x); assume x << 3.5 x 10-4 1.00 x 10-14 = (x)(3.5 x 10-4) x = 2.9 x 10-11 M = [H3O+]

Chapter 16

528

The concentrations are: [K+] = 3.45 x 10-2 M [HBrO] = [OH-] = 3.5 x 10-4 M [BrO-] = 3.42 x 10-2 M

[H3O+] = 2.9 x 10-11 M 16.45 This problem asks for the concentration of all ionic species present in the solution. Begin

by analyzing the chemistry. H2CO3 is a diprotic acid with equilibria: H2O + H2CO3 HCO3

- + H3O+ Ka1 = 4.5 x 10-7 H2O + HCO3

- CO3-2 + H3O+ Ka2 = 4.7 x 10-11

Every aqueous solution has the water equilibrium: H2O + H2O H3O+ + OH- Kw = 1.00 x 10-14

Thus, the ionic species present in the solution are: HCO3

-, CO3-2, H3O+, and OH-

Set up concentration tables, write equilibrium expressions, and solve for the ionic concentrations.

Reaction: H2O + H2CO3 HCO3- + H3O+

Start(10 –2 M) ---- 1.55 0 0

Change(10 –2 M) ---- -x +x +x

Final(10 –2 M) solvent 1.55-x x x

Ka1x

x−

== −−

2

27

10 x 55.101 x 5.4 ; assume x << 1.55 x 10-2

2

27

10 x 55.110 x 5.4 −

− = x;

from which x2 = 6.98 x 10-9 and x = 8.4 x 10-5; assumption valid

[H3O+] = [HCO3-] = 8.4 x 10-5 M

[H2CO3] = 1.55 x 10-2 M Now set up a concentration table for the second equilibrium:

Reaction: H2O + HCO3- CO3

-2 + H3O+

Start(10 –5 M) ---- 8.4 0 8.4

Change(10 –5 M) ---- -x +x +x

Chapter 16

529

Final(10 –5 M) solvent 8.4-x x 8.4 + x

Ka2xxx

−+== −

−−

5

511

10x 4.8)10x 4.8(

10 x 7.4 ; assume x << 8.4 x 10-5

x = 4.7 x 10-11M = [CO3

2-];

Now use a third concentration table to determine the ionic concentrations of hydronium and hydroxide ions:


Start(M) ---- ---- 8.4 x 10-5 0

Change (M) ---- ---- + x + x

Final (M) solvent solvent 8.4 x 10-5 + x x

Kw = 1.00 x 10-14 = (x)(8.4 x 10-5 + x); assume x << 8.4 x 10-5 1.00 x 10-14 = (x)(8.4 x 10-5) x = 1.2 x 10-10 M = [OH-]; assumption valid ionic concentrations: [H3O+] = [HCO3

-] = 8.4 x 10-5M [CO3

2-] = 4.7 x 10-11M [OH-] = 1.2 x 10-10 M

16.46 This problem asks for the concentration of all ionic species present in the solution. Begin by analyzing the chemistry. H2SO3 is a diprotic acid with equilibria:

H2O + H2SO3 HSO3- + H3O+ Ka1 = 1.4 x 10-2

H2O + HSO3- SO3

-2 + H3O+ Ka2 = 6.3 x 10-8 Every aqueous solution has the water equilibrium: H2O + H2O H3O+ + OH- Kw = 1.00 x 10-14

Thus, the ionic species present in the solution are: HSO3

-, SO3-2, H3O+, and OH-

Set up concentration tables, write equilibrium expressions, and solve for the ionic concentrations. Reaction: H2O + H2SO3 HSO3

- + H3O+

Start(M) ---- 0.355 0 0

Change(M) ---- -x +x +x

Chapter 16

530

Final(M) solvent 0.355 - x x x

Ka1x

x−

== −

355.010 x 4.1

22 ;

This must be solved using the quadratic equation: x = 0.064 x = 0.064 M = [H3O+] = [HSO3

-] [H2SO3] = 0.355 M – 0.064 M = 0.291 M

Use these values as initial concentrations in the concentration table for the second equilibrium reaction:

Reaction: H2O + HSO3

- SO3-2 + H3O+

Start(M) ---- 0.064 0 0.064


Final (M) solvent 0.064 - x x 0.064 - x


Ka2xxx

−+== −

064.0)064.0(01 x 3.6 8 ; assume x << 0.064

x = 6.3 x 10-8 M = [SO3

2-]; assumption valid Because the hydronium concentration is large the water equilibrium can be considered negligible:

[OH-] = 13--14

3

w 10 x 6.1.0640

10 x 00.1]O[H

==+K M

The ionic concentrations are: [H3O+] = [HSO3

-]= 0.064 M [SO3

2-]= 6.3 x 10-8M [OH-] = 1.6 x 10-13 M

16.47 There are two amine groups, one at either end of the molecule, each of which can accept

a proton from a water molecule. Convert the line structure to a Lewis structure using the standard procedures, then show the transfer of one proton to each N atom:

Chapter 16

531

16.48 First determine the Lewis structures following standard procedures, then show proton

transfer from the O–H group of acetic acid to the N atom of ammonia:

+ -

16.49 To determine the pH of a solution, follow the standard procedure for working equilibrium

problems: 1.) Major species are H2O, Na+, and F-; 2.) The dominant acid-base equilibrium is H2O + F- HF + OH-;

3.) Keq = Kb = a

w

KK ; from Table 16-3, Ka = 6.3 x 10-4, from which

Kb = 4

14

10 x 6.310 x 1.0

−

−

= 1.6 x 10-11;

Chapter 16

532

4.) Concentration table is:

Reaction: H2O + F- HF + OH-

Start (M) --- 0.250 0 0

Change (M) --- – x + x + x


Kb = 1.6 x 10-11 = x

x−

=250.0][F

][OH[HF] 2

eq-

eq-

eq ; Assume x << 0.250:

1.6 x 10-11 = 0.250



16.50 To determine the pH of a solution, follow the standard procedure for working equilibrium

problems: 1.) Major species are H2O, NH4+, and Cl-; 2.) The dominant acid-base equilibrium is H2O + NH4+ NH3 + H3O+;

3.) Keq = Ka; from Appendix E, Ka = 5-

-14

b

w

10 x 8.110 x 1.0=

KK = 5.6 x 10-10;

4.) The concentration table is:


Start (M) --- 0.025 0 0

Change (M) --- – x + x + x


Ka = 5.6 x 10-10 = x

x−

=025.0][NH

]O[H][NH 2

eq+4

eq+

3eq3 ; Assume x << 0.025:

5.6 x 10-10 = 0.025


[H3O+] = 3.7 x 10-6 M, and pH = -log(3.7 x 10-6) = 5.43.

16.51 To determine concentrations of species in a solution, follow the standard procedure:

Chapter 16

533

1.) This is a strong acid. Major species are H2O, H3O+, and HSO4-; 2.) The dominant acid-base equilibrium is H2O + HSO4- SO42- + H3O+; 3.) Keq = Ka2; from Table 16-2, Ka2 = 1.0 x 10-2; 4.) In this solution, there is a hydronium ion from the strong acid present initially:

Reaction: H2O + HSO4- SO42- + H3O+

Start (M) ---- 2.00 0 2.00

Change (M) ---- - x +x +x

Final (M) solvent 2.00 – x x 2.00 + x

Keq = 1.0 x 10-2 = xxx

−+=

00.2)00.2(

][HSO

]O[H][SO

eq-4

eq+

3eq-2

4 ; Assume x << 2.00:

1.0 x 10-2 = (2.00)

)(2.00)(x, from which x = 1.0 x 10-2; assumption valid

[H3O+] = 2.00 + 0.010 = 2.01 M; [SO42-] = 1.0 x 10-2 M; [HSO4-] = 2.00 – x = 1.99 M

16.52 To determine concentrations of species in a solution, follow the standard procedure: 1.) This is a salt. Major species are H2O, Na+, and NO2-; 2.) The dominant acid-base equilibrium is H2O + NO2- HNO2 + OH-;

3.) Keq = a

w

KK

= Kb; from Table 16-1, Ka = 5.6 x 10-4, from which

Kb = 4

14

10 x 5.610 x 1.0

−

−

= 1.8 x 10-11

4.)

Reaction: H2O + NO2- HNO2 + OH-

Start (M) ---- 0.200 0 0


Final (M) Solvent 0.200 – x x x

Keq = 1.8 x 10-11 = x

x−

=200.0][HNO

][OH][NO 2

eq2

eq-

eq-2 ; Assume x << 0.200:

1.8 x 10-11 = 0.200

)( 2x , from which x2 = 3.6 x 10-12; x = 1.9 x 10-6; assumption valid

[OH-] = [HNO2] = 1.9 x 10-6 M;

Chapter 16

534

[NO2- ] = 0.200 M

[H3O+] = 6

14w

10 x 1.910 x 1.0

][OH −

−

− =K= 5.3 x 10-9 M

16.53 (a) H2SO4 is a strong acid, so the major species in solution are H2O, HSO4-, and H3O+.

The hydrogensulfate ion is a weak acid, so the equilibrium reaction that determines pH is: H2O + HSO4- SO42- + H3O+;

(b) Na2SO4 is a salt, so the major species in solution are H2O, SO42-, and Na+. The sulfate ion is the conjugate base of a weak acid, so the equilibrium reaction that determines pH is:

SO42- + H2O HSO4- + OH-; (c) CO2 associates with H2O when it dissolves in water, so the major species in solution

are CO2, H2CO3, and H2O. Carbonic acid is a weak acid, so the equilibrium reaction that determines pH is:

H2O + H2CO3 HCO3- + H3O+; (d) NH4Cl is a salt, so the major species in solution are H2O, Cl-, and NH4+. The

ammonium ion is the conjugate acid of a weak base, so the equilibrium reaction that determines pH is:

H2O + NH4+ NH3 + H3O+. 16.54 (a) NH4NO3 is a salt, so the major species in solution are H2O, NO3-, and NH4+. The

ammonium ion is the conjugate acid of a weak base, so the equilibrium reaction that determines pH is: H2O + NH4+ NH3 + H3O+;

(b) KH2PO4 is a salt, so the major species in solution are H2O, H2PO4-, and K+. H2PO4- is the conjugate base of a weak acid and is also a weak acid itself, so there are two equilibrium reactions that might determines pH: H2O + H2PO4- HPO42- + H3O+, pKa2 = 7.21, and

H2O + H2PO4- H3PO4 + OH-, pKeq = pKw – pKa1 = 14.00 – 2.16 = 11.84; The equilibrium with smaller pK dominates: H2O + H2PO4- HPO42- + H3O+; (c) Na2O is a salt that dissolves to form O2-, an extremely strong base that reacts to

completion with water: O2- + H2O → 2 OH-, so the major species in solution are H2O, Na+, and OH-. There is no acid-base equilibrium to consider, so pH is determined by [OH-]initial;

(d) HCO2H is a weak acid, so the major species in solution are H2O and HCO2H. The equilibrium reaction that determines pH is: H2O + HCO2H HCO2- + H3O+.

16.55 Tabulated equilibrium constants for acid-base reactions always refer to reactions in which

H2O is one of the reactants. The reaction in this problem is the reverse of a base reaction:

HPO42- (aq) + OH- (aq) PO43- (aq) + H2O (l) Table 16-2 lists Ka values for phosphoric acid:

Chapter 16

535

HPO42- (aq) + H2O (l) PO43- (aq) + H3O+(aq) Ka3 = 4.8 x 10-13

Ka and Kb for a conjugate acid-base pair are related through Ka Kb = Kw:

Kb = 31

14

10 x 4.810 x 1.0

−

−

= 2.1 x 10-2;

Thus, Keq = b

1K = 48

16.56 Tabulated equilibrium constants for acid-base reactions always refer to reactions in which

H2O is one of the reactants. The reaction in this problem is the reverse of an acid reaction:

HPO42- (aq) + H3O+ (aq) H2PO4- (aq) + H2O (l) Table 16-2 lists Ka values for phosphoric acid:

H2PO4- (aq) + H2O (l) HPO42- (aq) + H3O+ (aq) Ka2 = 6.2 x 10-8

Keq = a

1K = 1.6 x 107

16.57 (a) An acid-base equilibrium reaction involves proton transfer, in this case from boric

acid to water:

(b) To calculate the pH of a solution, follow the standard procedure for equilibrium calculations:

Reaction: H2O + H3BO3 H2BO3- + H3O+

Start (M) ---- 0.050 0 0



Ka= 5.4 x 10-10 = x

x−

=050.0]BO[H

]O[H]BO[H 2

eq33

eq+

3eq-

23 ; Assume x << 0.050:

5.4 x 10-10 = 0.050

2x,

from which x2 = 2.7 x 10-11 and x = 5.2 x 10-6; assumption valid [H3O+] = 5.2 x 10-6 M, and

pH = -log(5.2 x 10-6) = 5.28

Chapter 16

536

16.58 (a) An acid-base equilibrium reaction involves proton transfer, in this case from water to

hydrazine:

(b) To calculate the pH of a solution, follow the standard procedure for equilibrium

calculations:

Reaction: H2O + N2H4 HN2H4+ + OH-

Start (M) ---- 0.200 0 0



Kb = 1.3 x 10-6 = x

x−

=200.0]H[N

][OH]H[HN 2

eq42

eq-

eq+

42 ; Assume x << 0.200:

1.3 x 10-6 = 0.200

2x,

from which x2 = 2.6 x 10-7 and x = 5.1 x 10-4; assumption valid [OH-] = 5.1 x 10-4 M,

pOH = -log(5.1 x 10-4) = 3.29, and pH = 14.00 – 3.29 = 10.71 16.59 There is much interesting chemical information provided in the statement of this

problem, but the calculation is a straightforward equilibrium determination for a solution of a weak base. Follow the standard procedures to determine the pH. Begin by constructing an equilibrium table, write the equilibrium expression, and solve for the concentration of hydroxide ions.

Reaction: H2O + LSD LSDH + + OH-

Start (M) ---- 0.55 0 0


Final (M) Solvent 0.55 - x x x


Kb = 7.6 x 10-7 = x

x−

=55.0[LSD]

][OH][LSDH 2

eq

eq-

eq+

; Assume x << 0.55:

Chapter 16

537

7.6 x 10-7 = 0.55



16.60 There is much interesting chemical information provided in the statement of this

problem, but the calculation is a straightforward equilibrium determination for a solution of a weak base. Follow the standard procedures to determine the pH. Begin by constructing an equilibrium table, write the equilibrium expression, and solve for the concentration of hydroxide ions.

Reaction: H2O + Morphine MorphineH + + OH-

Start (M) ---- 0.015 0 0



Kb = 7.9 x 10-7 = x

x−

=015.0[Morphine]

][OH][MorphineH 2

eq

eq-

eq+

; Assume x << 0.015

7.9 x 10-7 = 0.015


[OH-] = 1.1 x 10-4 M,

pOH = -log(1.1 x 10-4) = 3.96, and pH = 14.00 – 3.96 = 10.04. 16.61 To identify an acid from the pH of its solution, use the pH to calculate the equilibrium

constant of the acid. For pH = 2.71, [H3O+]eq = 1.95 x 10-3 M = [A-]; [HA]eq = (0.060 M – 0.00195 M) = 0.058 M;

Ka =

[A- ]eq [H3O+ ]eq

[HA]eq =

(0.058))10 x (1.95 2-3

= 6.6 x 10-5; pKa = 4.18;

The acid is benzoic acid, listed in Appendix E , pKa = 4.19. 16.62 Follow the standard procedure for dealing with equilibrium calculations: The major species are Na+, A-, and H2O, and the acid-base equilibrium is: A-(aq) + H2O (l) HA (aq) + OH- (aq) Set up a concentration table and use it to determine Keq: pOH = 14.00 – pH = 3.00;

[OH-] = 10-3.00 = 1.00 x 10-3 M;

Chapter 16

538

Reaction: H2O + A- HA + OH-

Start (M) ---- 0.0100 0 0

Change (M) ---- -0.00100 +0.00100 +0.00100

Final (M) solvent 0.0090 0.00100 0.00100


Keq = Kb = eq

-eq

-eq

][A][OH[HA] =

(0.0090))10 x (1.00 2-3

= 1.1 x 10-4;

Ka = 4-

-14

b

w

10 x 1.110 x 00.1=

KK

= 9.1 x 10-11

16.63 Molecular pictures must show the correct relative numbers of the various species in the

solution. From the starting condition (six molecules of oxalic acid), make appropriate changes and then draw new pictures:

(a) Hydroxide ions react with oxalic acid to form water and hydrogen oxalate ions: H2C2O4 + OH- H2O + HC2O4- The picture shows 2 molecules of oxalic acid and

4 each of water and hydrogenoxalate:

(b) When all oxalic acid has reacted, hydroxide ions react with hydrogen oxalate ions to form water and oxalate ions:

HC2O4- + OH- H2O + C2O42-. The picture shows 4 hydrogen oxalate ions, 8 water molecules, and 2 oxalate ions:

Chapter 16

539

(c) NH3, a weak base, accepts a proton from oxalic acid, a weak acid: H2C2O4 + NH3 NH4+ + HC2O4- The picture shows 2 oxalic acid molecules and four each ammonium and

hydrogenoxalate ions:

16.64 The acidic or basic nature of a solution depends on the dominant equilibrium, which is

determined by the major species present: (a) CH3CO2H is a weak acid, major species are H2O and CH3CO2H; dominant

equilibrium is CH3CO2H + H2O CH3CO2- + H3O+, and the solution is acidic; (b) NH3 is a weak base, major species are H2O and NH3; dominant equilibrium is NH3 + H2O NH4+ + OH-, and the solution is basic; (c) NH4Cl is a salt, major species are H2O, NH4+, and Cl-; dominant equilibrium is NH4+ + H2O NH3 + H3O+, and the solution is acidic; (d) NaCH3CO2 is a salt, major species are H2O, CH3CO2-, and Na+; dominant

equilibrium is CH3CO2- + H2O CH3CO2H + OH-, and the solution is basic;

Chapter 16

540

(e) NH4CH3CO2 is a salt, major species are H2O, CH3CO2-, and NH4+; dominant equilibrium is CH3CO2- + NH4+ CH3CO2H + NH3, for which Ka = Kb; thus this solution is neutral.

16.65 The chemical reaction that occurs is:

P4O10 + 6 H2O → 4 H3PO4 (a) The major species present are H2O, and H3PO4 (b) The minor species are (in order of highest concentration to lowest):

H2PO4-, H3O+, HPO4

2-, OH- , and PO43-

(c) The dominant equilibrium that determines the pH is: H3PO4 + H2O H2PO4

- + H3O+ Set up a concentration table, solve for hydronium ion concentration, then calculate the pH. Determine the initial concentration using standard stoichiometric procedures:

n(H3PO4) = 3.5 g P4O10

104

43

OP mol 1POH mol 4

g 283.88mol 1

= 0.0493 mol;

[H3PO4] = L 1.50mol 0.0493 = 0.033 M

Reaction: H2O + H3PO4 H2PO4- + H3O+

Start (M) ---- 0.033 0 0

Change (M) ---- -x + x + x


Now substitute into the equilibrium constant expression and solve for x:

Keq = 0.0069 =eq43

eq3eq-42

]PO[H]O[H]PO[H +

= -x.

x0330

2

; solve by the quadratic equation

[H3O+] = 0.012 M, pH = -log(0.012) = 1.92

16.66 (a) The salt generates Na+ and SO32-, CH3CO2H is a weak acid, and H2O is always a

major species in aqueous solution; (b and c) CH3CO2H is the acid, SO32- is the base, HSO3- is the conjugate acid, and

CH3CO2- is the conjugate base:

Chapter 16

541

16.67 Proton transfer occurs from the carboxylic acid O–H (shown screened below) and the

amino nitrogen atom:

16.68 To determine the mass percent of vinegar we must first determine how much of each

species is present in the solution using the standard procedure: 1.) This is a weak acid. Major species are H2O and CH3CO2H

2.) The dominant acid-base equilibrium is CH3CO2H + H2O CH3CO2- + H3O+;

3.) Keq = Ka = 1.8 x 10-5

4.) In this solution, the pH corresponds to the equilibrium concentrations of hydronium ions and acetate ions:

[H3O+] = [CH3CO2- ] = 10-2.39 = 0.00407 M

Assuming that the concentration of CH3CO2H is much larger than 0.00407 M:

Ka =1.8 x 10-5 = x

2

23

233 )00407.0(H]CO[CH

]CO][CHO[H =−+

x = 0.92 M = [HCH3CO2]; assumption valid

m(acetic acid) =

mol 1g 60.05

L 1mol 0.92

=55 g/L or 0.055 g/mL acetic acid

mass % = %10007.1055.0

sample ofdensity acid acetic ofdensity ×= = 5.1%

16.69 Net ionic equations show only the reacting species. Remember that strong acids generate

H3O+ in solution and react to completion with weak bases, and strong bases generate OH- in solution and react to completion with weak acids:

(a) strong base reacting with weak acid: OH- + C6H5CO2H H2O + C6H5CO2- ; (b) strong acid reacting with weak base: H3O+ + (CH3)3N H2O + (CH3)3NH+;

Chapter 16

542

(c) weak base reacting with weak acid: SO42- + CH3CO2H HSO4- + CH3CO2- HSO4-, pKa = 1.99; CH3CO2H, pKa = 4.75; HSO4- is stronger, so this reaction proceeds

to a small extent; (d) strong base reacting with weak acid: OH- + NH4+ H2O + NH3; (e) weak base reacting with weak acid: HPO42- + NH3 PO43- + NH4+ ; HPO42-, pKa = 12.32; NH4+, pKa = 9.25; NH4+ is stronger, so this reaction proceeds to

a small extent. 16.70 (a) H2CO3 is acidic. H2CO3 + H2O HCO3

− + H3O+

(b) KHCO3 is both. HCO3− + H2O CO3

2− + H3O+ (acting as acid) HCO3

− + H2O H2CO3 + OH- (acting as base)

(c) NH3 is basic. NH3 + H2O NH4+ + OH-

(d) NaCl is neither. (e) Na2SO4 is basic. SO4

2 − + H2O HSO4− + OH-

(f) SO2 is acidic. SO2 + H2O H2SO3 H2SO3 + H2O HSO3

–+ H3O+

SO2 + 2 H2O HSO3–+ H3O+

(g) Li2O is basic. Li2O + H2O 2 Li+ (aq) + 2 OH- (aq) 16.71 Follow the standard procedure for dealing with equilibrium calculations: (a) The major species are Na+, HCO3-, and H2O; There are two equilibria involving major species: HCO3- (aq) + H2O (l) CO32- (aq) + H3O+ (aq) pKeq = pKa2 = 10.33 HCO3- (aq) + H2O (l) H2CO3 (aq) + OH- (aq)

pKeq = pKw – pKa1 =14.00 – 6.35 = 7.65;

The equilibrium with the larger Keq (smaller pKeq) dominates, making this solution basic; (b) Set up a concentration table, solve for hydroxide ion concentration, then calculate the

pH. Determine the initial concentration using standard stoichiometric procedures:

[HCO3-] = L 0.150mol 0.0228=

Vn = 0.152 M;

Reaction: H2O + HCO3- H2CO3 + OH-

Start (M) ---- 0.152 0 0

Change (M) ---- -x + x + x


Now substitute into the equilibrium constant expression and solve for x:

Chapter 16

543

Keq = 10-7.65 = 2.2 x 10-8;

2.2 x 10-8 =

[H2CO3]eq[OH- ]eq

[HCO3- ]eq

= -x.

x1520

2

; assume x << 0.152

x2 = 3.3 x 10-9 , from which x = 5.8 x 10-5; assumption valid [OH-] = 5.8 x 10-5 M,

pOH = -log(5.8 x 10-5) = 4.24 and pH = 14.00 – 4.24 = 9.76

16.72 (a) CH3CO2- + H2O CH3CO2H + OH-

a

w

KK

NH4+ + H2O NH3 + H3O+

b

w

KK

H3O+ + OH- 2 H2O w

1K

NH4+ + CH3CO2

- NH3 + CH3CO2H

Keq = ba

w

KKK

(b) to determine all proton transfer reactions, first determine all major acid/base species

in the solution. From part (a) we can see that the major species are: Bases: CH3CO2

-, NH3, H2O, and OH-. Acids: CH3CO2H, NH4

+, H3O+, and H2O. Now make a list of the reactions between each acid/base pair CH3CO2

- + H2O CH3CO2H + OH- NH4

+ + H2O NH3 + H3O+ H3O+ + OH- 2 H2O

NH3 + H2O NH4+ + OH-

CH3CO2H + H2O CH3CO2- + H3O+

NH4+ + CH3CO2

- NH3 + CH3CO2H CH3CO2

- + H3O+ CH3CO2H + H2O NH3

+ H3O+ NH4+ + H2O

CH3CO2H + OH- CH3CO2- + H2O

NH4+ + OH- NH3 + H2O

(c) Using the equations from part a, set up an equilibrium table, write the Keq expression

and solve for the desired concentrations:

Chapter 16

544

Reaction: CH3CO2- + H2O CH3CO2H + OH-

Start (M) 0.25 ---- 0 0

Change (M) -x ---- + x + x

Final (M) 0.25 - x solvent x x

Kb = x

xKK

−=== −−

−

25.0)(

]CO[CHH]CO][CH[OH

10 x 1.810 x 1.00 2

23

23-

10

14

a

w ; assume x << 0.25

5.6 x 10-5 =25.0

2x

x = 1.2 x 10-5; assumption correct [OH-] = 1.2 x 10-5 M Repeat with ammonium to determine the concentration of hydronium ions:

Reaction: NH4+ + H2O NH3 + H3O+

Start (M) 0.25 ---- 0 0

Change (M) -x ---- + x + x

Final (M) 0.25 - x solvent x x

Kb = x

xKK

−=== +

+

−

−

25.0)(

][NH]][NHO[H

10 x 1.810 x 1.00 2

4

3310

14

b

w ; assume x << 0.25

5.6 x 10-5 =25.0

2x

x = 1.2 x 10-5; assumption correct [H3O+] = 1.2 x 10-5 M

(d) Since the concentrations of hydroxide and hydronium ions are equal, we can expect both to react completely with each other to form water. The resulting equilibrium is that of water which has a pH of 7.0

H3O+ + OH- 2 H2O

16.73 Compare Ka values to determine the extent of the reaction.

(a) H2S is the stronger acid : H2S + NH3 HS- + NH4+;

(b) HSO4- is the stronger acid, therefore the reaction is significant:

C2H5NH2 + HSO4- C2H5NH3

+ +SO42-;

(c) HCN is a weaker acid than C5H5NH+, therefore the reaction proceeds only to a small extent : C5H5N + HCN C5H5NH+ + CN-;

(d) ammonium is the stronger acid: NH4+ + PO4

3- NH3 + HPO42-

(e) HClO is the slightly weaker acid:

Chapter 16

545

HClO(aq) + HONH2(aq) ClO-(aq) + HONH3+(aq)

16.74 (a) HBr is a strong acid, so when this gas bubbles through water it generates hydronium

ions. The major species are H2O, H3O+, Br-, Ca2+, and OH-, and the reaction that goes to completion is H3O+ + OH- → 2 H2O;

(b) The major species are H2O, Na+, HSO4-, and OH-, and the reaction that goes to completion is HSO4- + OH- → H2O + SO42-;

(c) The major species are H2O, NH4+, I-, Pb2+, and NO3-, and the reaction that goes to completion is formation of PbI2 precipitate, Pb2+ + 2 I- → PbI2(s).

16.75 (a) H2SO4(l) + H2SO4(l) H3SO4

+(solv) + HSO4-(solv);

Either doubly bonded oxygen atom can accept a proton.

(c) H2SO4(l) + HClO4(l) H3SO4

+(solv) + ClO4-(solv);

Documents

Chapter 16 16.1 HBr is a strong acid and will …Chapter 16 506 16.1 HBr is a strong acid and will therefore give its hydrogen atom to a water molecule, making a hydronium cation and