acid base problems solutions

7/23/2019 Acid Base Problems Solutions

1/20

Acid-Base Equilibrium Problems

1. Write equations which represent the dissociation of each of these acids or bases in aqueoussolution. Use a single arrow in the case of a strong acid or base, and a double arrow to representthe equilibrium condition that exists in the solution of a weak acid or base. Show each step ofdissociation for polyprotic acids.a) KOH

KOH(aq)K+(aq)+ OH

-(aq)

b) H3AsO4Step 1: H3AsO4 (aq)+ H2O(l)H2AsO4

-(aq) + H3O

+(aq)

Step 2: H2AsO4-(aq) + H2O(l)HAsO4

2-(aq) + H3O

+(aq)

Step 3: HAsO42-

(aq) + H2O(l)AsO43-

(aq) + H3O+(aq)

c) HClO4

HClO4 (aq)+ H2O(l)ClO4-(aq) + H3O

+(aq)

d) HCN(aq)

HCN(aq)+ H2O(l)CN-(aq) + H3O

+(aq)

e) C6H5NH2(a weak base)

C6H5NH2 (aq) + H2O(l)C6H5NH3+(aq) + OH

-(aq)

2. Benzoic acid, HC6H5CO2, is an organic acid whose sodium salt, NaC6H5CO2, has long been used as asafe food additive to protect beverages and many foods against harmful yeasts and bacteria. Theacid is monoprotic. Write the Kaexpression for this acid.

HC6H5CO2 (aq)+ H2O(l)C6H5CO2-(aq) + H3O

+(aq)

][

]][[

266

3266

COHHC

OHCOHCK

a

3.

Write the equilibrium equations and the Kbexpressions for each of the following bases.a) CN-(cyanide ion)

CN-(aq) + H2O(l)HCN(aq)+ OH-(aq)

][

]][[

HCN

OHHCNK

b

b) C2H3O2-(acetate ion)

C2H3O2-(aq) + H2O(l)HC2H3O2(aq)+ OH

-(aq)

][

]][[

232

232

OHC

OHOHHCK

b

c) C6H5NH2(aniline)

C6H5NH2 (aq)+ H2O(l)C6H5NH3+(aq)+ OH

-(aq)

][

]][[

256

356

NHHC

OHNHHCK

b

d) H2O(l)+ H2O(l)H3O+(aq)+ OH

-(aq)

Kb=[H3O+

][OH-

]


2/20

4.

Find the pH and % ionization of a 0.065 M solution of formic acid, (Ka=1.8 x 10-4). [2.47, 5.2%]

HCHO2 (aq)+ H2O(l)CHO2-(aq) + H3O

+(aq) Ka=1.8 x 10

-4

MR 1 1 1 1

I 0.065 --- 0 0

C -x --- +x +x

E 0.065-x ---- x x

0102.1108.1

108.1102.1

)065.0(108.1

][

]][[

542

245

24

2

32

xx

xx

x

x

HCHO

OHCHOK

a

a= 1, b=1.8x10-4, c=-1.2x10-5x1=-3.6x10

-3, x2=3.4x10-3

[H3O+]=3.4x10-3M

pH = -log[H3O+]pH=2.47

% ionization= [H3O+]/[Acid]initialx100%

= (3.4x10-3)/0.065 x100%= 5.2%

5. Find the pH of a 0.325 M acetic acid solution, given the Kais 1.8 x 10-5. [2.62]

HC2H3O2 (aq)+ H2O(l)C2H3O2-(aq) + H3O

+(aq) Ka=1.8 x 10

-5

MR 1 1 1 1I 0.325 --- 0 0

C -x --- +x +x

E 0.325-x ---- x x

3

26

25

232

3232

104.2

109.5

)325.0(108.1

][

]][[

x

x

x

x

OHHC

OHOHCK

a

[H3O+]=2.4x10-3M

pH = -log[H3O+]

pH=2.62

% ionization= [H3O+]/[Acid]initialx100%

= (2.4x10-3)/0.325 x100%

= 0.74 %

6. Find the pH of a solution that contains 0.0034 M lactic acid(HC3H5O3) (Ka=1.4 x 10-4).


+(aq) Ka=1.4 x 10

-4

MR 1 1 1 1

I 0.0034 --- 0 0

C -x --- +x +x

E 0.0034-x ---- x x

0108.4104.1

)0034.0(104.1

][

]][[

742

24

353

3353

xx

x

x

OHHC

OHOHCK

a

a= 1, b=1.4x10-4, c= -4.8x10-7x1=6.3x10

-4, x2= -7.7x10-4

[H3O+]=6.3x10-4M

pH = -log[H3O+]

pH=3.20

0.065/Ka< 1000

approx will not work

0.325/Ka> 1000

approx will work

0.0034/Ka< 1000



3/20

7.

Find the hydronium ion concentration and pH for a 0.056 M solution of propanoic acid (Ka= 1.4 x10-5). [ H3O

+=8.9x10-4 pH=3.05]


+(aq) Ka=1.4 x 10

-5

MR 1 1 1 1

I 0.056 --- 0 0

C -x --- +x +x

E 0.056-x ---- x x

4

25

253

3253

109.8

)056.0(104.1

][

]][[

x

x

x

OHHC

OHOHCK

a

[H3O+]=8.9x10-4M

pH = -log[H3O+]

pH=3.05

8. Find the pH of a 0.600 M solution of methylamine CH3NH2. Kb= 4.4 x 104. [12.20]

CH3NH2 (aq)+ H2O(l)CH3NH3+(aq) + OH

-(aq) Kb=4.4 x 10

-4

MR 1 1 1 1

I 0.60 --- 0 0

C -x --- +x +x

E 0.60-x ---- x x

2

24

23

33

106.1

)60.0(104.4

][

]][[

x

x

x

NHCH

OHNHCHK

b

[OH-]=1.6x10-2MpOH = -log[OH-]pOH = 1.80

pH=12.20

9. What is the pH of a 5.6x10-4M butanoic acid solution, given the pKa= 4.82. [4.08]


+(aq)

MR 1 1 1 1

I 5.6x10-4 --- 0 0

C -x --- +x +x

E 5.6x10-4-x ---- x x

pKa=4.82Ka=10

-4.82Ka=1.5 x 10

-5

0104.8105.1

)106.5(105.1

][

]][[

952

4

25

274

3274

xx

x

x

OHHC

OHOHCK

a

a= 1, b=1.5x10-5, c= -8.4x10-9x1=8.4x10

-5, x2= -9.9x10-5

[H3O+]=8.4x10-5M

pH = -log[H3O+]

pH=4.08

0.056/Ka> 1000

approx will work

0.60/Ka> 1000

approx will work

5.6x10-4/Ka> 1000

approx will work


4/20

10.

Calculate the [H3O+], the pH and the % ionization for a 0.50 mol/L HCN solution. [1.8x10-5,4.74,

3.6x10-3 %]

HCN(aq)+ H2O(l)CN-(aq) + H3O

+(aq)

MR 1 1 1 1

I 0.50 --- 0 0

C -x --- +x +x

E 0.50-x ---- x x

5

210

3

108.1

)50.0(102.6

][

]][[

x

x

x

HCN

OHCNK

a

[H3O+]=1.8x10-5M

pH = -log[H3O+]

pH=4.74

11.Calculate the [OH-], the pH and the % ionization for 0.25 mol/L ammonia solution. [2.1x10-3,11.32, 0.84%]

NH3 (aq)+ H2O(l)NH4+(aq) + OH

-(aq) Kb=1.8 x 10

-5

MR 1 1 1 1

I 0.25 --- 0 0

C -x --- +x +x

E 0.25-x ---- x x

3

25

3

4

101.2

)25.0(108.1

][

]][[

x

x

x

NH

OHNHK

b


pH=11.32

% ionization= [OH-]/[Base]initialx100%

= (2.1x10

-3

)/0.25 x100%= 0.84 %

12.Hydrazine, N2H4, has been used as a rocket fuel. Like ammonia, it is a weak base. A 0.15 Msolution has a pH of 10.70. What is the Kband pKbfor hydrazine and the pKaof its conjugate acid?[Kb=1.7x10

-6, pKb=5.77, pKaConj=8.23]

pH=10.70 pOH = 3.30 [OH-]=10-3.30 [OH-]= 5.0x10-4

N2H4 (aq)+ H2O(l)N2H5+(aq) + OH

-(aq) Kb= ?

MR 1 1 1 1

I 0.15 --- 0 0C -x --- +x +x

E 0.15-x=0.15

---- x=5.0x10-4

x=5.0x10-4

6

24

42

52

107.1

)15.0(

)100.5(

][

]][[

b

b

b

K

K

HN

OHHNK

pKb=-logKbpKb=5.77pKa Conj=14- pKbpKaConj=8.23

0.50/Ka> 1000

approx will work

0.25/Ka> 1000

approx will work


5/20

13.If the pH of a weak base solution is 9.5 and the original concentration of base was 0.30 M what is

the pOH, the concentration of OH-, the equilibrium concentration of the base and the Kbof the

base? [pOH = 4.50, [OH-]= 3.2x10-5, [B]=0.30 M, Kb=3.4x10-9]

pH=9.50 pOH = 4.50 [OH-]=10-4.50 [OH-]= 3.2x10-5

B(aq)+ H2O(l) BH+(aq) + OH

-(aq) Kb= ?

MR 1 1 1 1

I 0.30 --- 0 0C -x --- +x +x

E 0.30-x=0.30

---- x=3.2x10-5

x=3.2x10-5

14.

If the pH of a hypochlorous acid solution is 4.17, what is the initial concentration? [0.16 M]

pH=4.17 [H3O+]=10-4.17 [H3O

+]=6.8x10-5

HClO(aq)+ H2O(l)ClO-(aq) + H3O

+(aq) Ka=2.9x10

-8

MR 1 1 1 1

I I --- 0 0C -x --- +x +x

E I-x= I-6.8x10-5

---- x=6.8x10-5

x=6.8x10-5

15.

If the pH of a solution is equal to 8.70 and the Kb=9.6x10-7what was the original concentration of

the base? [3.1x10-5M]

pH=8.70 pOH = 5.30 [OH-]=10-5.30 [OH-]= 5.0x10-6

B(aq)+ H2O(l) BH+(aq) + OH-(aq) Kb= 9.6x10-7MR 1 1 1 1

I I --- 0 0

C -x --- +x +x

E I-x=I-5.0x10-6

---- x=5.0x10-6

x=5.0x10-6 5

1127

6

267

101.3

105.2108.4106.9

)100.5(

)100.5(106.9

][

]][[

I

I

I

B

OHBHK

b

16.0

106.4100.2109.2

)108.6(

)108.6(109.2

][

]][[

9128

5

258

3

I

I

I

HClO

OHClOK

a

9

25

104.3

)30.0(

)102.3(

][

]][[

b

b

b

K

K

B

OHBHK


6/20

16.

A nitrous acid solution is 1.34% ionized at equilibrium. What is the hydronium ion concentration,pH and initial acid concentration? [ [H3O

+]=0.0531 M, pH=1.274, [HNO2]initial=3.96 M ]

HNO2 (aq)+ H2O(l)NO2-(aq) + H3O

+(aq) Ka=7.2x10

-4

MR 1 1 1 1

I I --- 0 0

C -x --- +x +x

E I-x=0.9866I

---- x=0.0134I

x=0.0134I

Since the acid is 1.34% ionized atequilibrium,

x=0.0134I

MI

I

I

I

I

I

HNO

OHNOK

a

96.3

1082.1102.7

9866.0

1080.1102.7

)9866.0(

)0134.0(102.7

][

]][[

44

244

24

2

32

x=0.0134Ix=(0.0134)(3.96)x=0.0531 M[H3O

+]=0.0531 MpH = -log[H3O

+]pH=1.274

17.

A solution of ethylamine is 1.25% ionized at equilibrium. What is the hydroxide ion concentrationand initial base concentration? [[OH-]=0.0340 M, [CH3CH2NH2]initial=2.72 M]

CH3CH2NH2 (aq)+ H2O(l)CH3CH2NH3+(aq) + OH

-(aq) Kb=4.3 x 10

-4

MR 1 1 1 1

I I --- 0 0

C -x --- +x +x

E I-x=0.9875I

---- x=0.0125I

x=0.0125I

Since the base is 1.25% ionized atequilibrium,

x=0.0125I

MI

I

I

I

I

I

NHCHCH

OHNHCHCHK

b

72.2

1058.1103.4

9875.0

1056.1103.4

9875.0

)0125.0(103.4

][

]][[

44

244

24

223

323

x=0.0125Ix=(0.0125)(2.72)x=0.0340 M[OH-]=0.0340 M


7/20

18.

Calculate the pH of the following solutions,a) 0.36 M sodium acetate solution [9.15]

C2H3O2-(aq) + H2O(l) HC2H3O2(aq)+ OH

-(aq) Kb=Kw/Ka Acid

MR 1 1 1 1

I 0.36 --- 0 0

C -x --- +x +x

E 0.36-x ---- X x

5

210

2

232

232

104.1

36.0106.5

36.0

][

]][[

x

x

x

x

K

K

OHC

OHOHHCK

aAcid

W

b

[OH-]=1.4x10-5M

pOH = -log[OH-]pOH = 4.85

pH=9.15

b) 0.25 M ammonium nitrate [4.92]

NH4+(aq)+ H2O(l)NH3 (aq) + H3O

+(aq) Ka=Kw/Kb Base

MR 1 1 1 1

I 0.25 --- 0 0

C -x --- +x +x

E 0.25-x ---- x x

5

210

2

3

33

102.125.0

106.5

25.0

][

]][[

x

x

x

x

K

K

NH

OHNHK

bBase

W

a

[H3O+]=1.2x10-5M

pH = -log[H3O+]

pH=4.92

c) 0.16 M calcium formate (Ca(CHO2)2) solution [8.48]

CHO2-(aq) + H2O(l) HCHO2(aq)+ OH

-(aq) Kb=Kw/Ka Acid

MR 1 1 1 1

I 0.32 --- 0 0

C -x --- +x +x

E 0.32-x ---- x X

6

211

2

2

2

102.4

32.0106.5

32.0

][

]][[

x

x

x

x

K

K

CHO

OHHCHO

K

aAcid

W

b

[OH-

]=4.2x10-6

MpOH = -log[OH-]pOH = 5.38

pH= 8.62

0.36/Kb> 1000

approx will work

0.25/Ka> 1000approx will work

0.16/Kb> 1000

approx will work


8/20

d) 0.45 M potassium hypochlorite [10.59]

ClO-(aq) + H2O(l) HClO (aq)+ OH-(aq) Kb=Kw/Ka Acid

MR 1 1 1 1

I 0.45 --- 0 0

C -x --- +x +x

E 0.45-x ---- x x

4

27

2

109.3

45.0104.3

45.0

][

]][[

x

x

x

x

K

K

ClO

OHHClOK

aAcid

W

b


pH= 10.59

19.

A 27.0 mL sample of 0.28 M acetic acid is titrated with 0.15 M calcium hydroxide.a)

Calculate the volume of calcium hydroxide required to reach the equivalence point. [25mL]b) Calculate the pH at equivalence. [8.96]

2 HC2H3O2 (aq)+ Ca(OH)2 (aq)Ca(C2H3O2)2 (aq)+ 2 H2O (l)c 0.28 0.15 0.073 ([C2H3O2

-]=0.15 M)V 0.027 0.025 0.052n 7.6x10-3 3.8x10-3 3.8x10-3

C2H3O2-(aq) + H2O(l) HC2H3O2(aq) + OH

-(aq)

MR 1 1 1 1I 0.15 -- 0 0

C -x --- x x

E 0.15-x -- x x

6

210

2

232

232

102.9

15.0106.5

15.0

][

]][[

x

x

x

x

K

K

OHC

OHOHHCK

aAcid

W

b

[OH-]=9.2x10-6pOH = -log[OH-]pOH=5.04pH=8.96

0.45/Kb> 1000

approx will work

0.15/Kb> 1000

approx will work


9/20

20.

A 175 mL sample of 0.30 M ammonia is titrated with 0.45 M hydrochloric acid.a) Calculate the volume of hydrochloric acid required to reach the equivalence point. [120mL]b) Calculate the pH at equivalence. [5.00]

HCl(aq)+ NH3 (aq) NH4Cl(aq)c 0.45 0.30 0.18 ([NH4

+]=0.18 M)V 0.12 0.175 0.295n 0.053 0.053 0.053

NH4+

(aq) + H2O(l)

NH3(aq) + H3O

+

(aq)MR 1 1 1 1

I 0.18 -- 0 0

C -x --- x x

E 0.18-x -- x x

5

210

2

3

33

100.1

18.0106.5

18.0

][

]][[

x

x

x

x

K

K

NH

OHNHK

bBase

W

a

[H3O+]=1.0x10-5

pH=-log[H3O+]

pH=5.00

21.

A 225 mL sample of 0.24 M formic acid is titrated with 0.30 M sodium hydroxide.a)

Calculate the volume of sodium hydroxide required to reach equivalence. [180mL]b) Calculate the pH at equivalence. [8.43]

HCHO2 (aq)+ NaOH(aq)

NaCHO2 (aq)+ H2O (l)c 0.24 0.30 0.13 ([C2H3O2-]=0.13 M]

V 0.225 0.180 0.405n 0.054 0.054 0.054

CHO2-(aq) + H2O(l) HCHO2(aq) + OH

-(aq)

MR 1 1 1 1

I 0.13 -- 0 0

C -x --- x x

E 0.13-x -- x x

6

211

22

2

107.2

13.0106.5

13.0

][

]][[

x

x

x

x

K

K

CHO

OHHCHOK

aAcid

W

b

[OH-]=2.7x10-6pOH = -log[OH-]pOH =5.57pH=8.43

0.18/Ka> 1000

approx will work

0.13/Kb> 1000

approx will work


10/20

22.A 180 mL sample of 0.16 M hydrazine is titrated with 0.12 M nitric acid.

a) Calculate the volume of nitric acid required to reach the equivalence point. [240mL]b)

Calculate the pH at equivalence. [4.70]

HNO3 (aq)+ N2H4 (aq) N2H5NO3 (aq)c 0.12 0.16 0.069 ([N2H5

+]=0.069 M)V 0.24 0.180 0.42n 0.029 0.029 0.029

N2H5+(aq) + H2O(l) N2H4 (aq) + H3O

+(aq)

MR 1 1 1 1

I 0.069 -- 0 0

C -x --- x x

E 0.069-x -- x x

5

29

2

52

342

100.2

069.0109.5

069.0

][

]][[

x

x

x

x

K

K

HN

OHHNK

bBase

W

a

[H3O+]=2.0x10-5

pH=4.70

23.

A 375 mL sample of 0.25 M hydrofluoric acid is titrated with 0.15 M potassium hydroxide.a) the volume of potassium hydroxide needed to reach the equivalence point [630 mL]b) the pH at equivalence. [8.07]

HF(aq)+ KOH(aq) KF(aq)+ H2O (l)c 0.25 0.15 0.094([F-]=0.094 M]V 0.375 0.63 1.005n 0.094 0.094 0.094

F-(aq) + H2O(l) HF(aq) + OH-(aq) Ka HF=6.6x10

-4

MR 1 1 1 1

I 0.094 -- 0 0

C -x --- x x

E 0.094-x -- x x

6

211

2

102.1

094.0105.1

094.0

][

]][[

x

x

x

x

K

K

F

OHHFK

aAcid

W

b

[OH-]=1.2x10-6pOH = -log[OH-]pOH =5.92pH=8.08

0.069/Ka> 1000

approx will work

0.094/Kb> 1000

approx will work


11/20

24.If 125 mL of 0.15 M hydrochloric acid is needed for 95 mL of ammonia to reach equivalence, find:a)

the concentration of the original ammonia solution. [0.20 M]b)

the pH of the solution at the equivalence point. [5.16]

HCl(aq)+ NH3 (aq) NH4Cl(aq)c 0.15 0.20 0.086 ([NH4

+]=0.086 M)V 0.125 0.095 0.22n 0.019 0.019 0.019

NH4+(aq) + H2O(l) NH3(aq) + H3O

+(aq) Kb ammonia=1.8x10

-5

MR 1 1 1 1I 0.086 -- 0 0

C -x --- x x

E 0.086-x -- x x

086.0106.5

086.0

][

]][[

210

2

3

33

x

x

x

K

K

NH

OHNHK

bBase

W

a

x=6.9x10-6[H3O

+]=6.9x10-6pH=-log[H3O

+]pH=5.16

25.

If 230 mL of 0.35 M cyanic acid (Ka=3.5x10-4) is titrated with 0.25 M calcium hydroxide. Calculate:

a) the pH of the acid sample before titration. [1.96]b)

the volume of calcium hydroxide needed to reach the equivalence point. [160 mL]c)


a)

HCNO(aq)+ H2O(l) CNO-(aq) + H3O

+(aq) Ka =3.5x10

-4

MR 1 1 1 1

I 0.35 --- 0 0

C -x --- +x +x

E 0.35-x ---- x x

x=1.1x10-2[H3O

+]=1.1x10-2pH=1.96

b) and c) 2 HCNO(aq)+ Ca(OH)2 (aq)Ca(CNO)2 (aq)+ 2 H2O (l)c 0.35 0.25 0.10 ([CNO-]=0.20 M)V 0.23 0.16 0.39n 0.081 0.0405 0.0405

CNO-(aq) + H2O(l) HCNO(aq) + OH-(aq)

MR 1 1 1 1I 0.20 -- 0 0

C -x --- x x

E 0.20-x -- x x

20.0109.2

20.0

][

]][[

211

2

232

232

x

x

x

K

K

OHC

OHOHHCK

aAcid

W

b

x=2.4x10-6[OH-]=2.4x10-6pOH = -log[OH-]pOH=5.62pH=8.38

0.086/Ka> 1000

approx will work

0.20/Kb> 1000

approx will work

0.35/Ka= 1000

approx will work

)35.0(105.3

][

]][[

24

3

x

x

HCNO

OHCNOK

a


12/20

26.A 65 mL sample of 0.45 M methylamine is titrated with 0.45 M nitric acid. Calculate:

a) the pH of the base sample before titration. [12.15]b)

the volume of nitric acid needed to reach the equivalence point. [65 mL]c)


a)

pH before titration

CH3NH2 (aq)+ H2O(l) CH3NH3+(aq) + OH

-(aq) Kb=4.4 x 10

-4

MR 1 1 1 1

I 0.45 --- 0 0C -x --- +x +x

E 0.45-x ---- x x

)45.0(104.4

][

]][[

24

23

33

x

x

NHCH

OHNHCHK

b

x=1.4x10-2[OH-]=1.4x10-2MpOH = -log[OH-]pOH = 1.85

pH=12.15

b) and c) HNO3 (aq)+ CH3NH2 (aq) CH3NH3NO3 (aq)+ 2 H2O (l)c 0.45 0.45 0.22 ([CH3NH3

+]=0.22 M)

V 0.65 0.65 0.13n 0.029 0.029 0.029

CH3NH3+ + H2O(l) CH3NH2 (aq)+ H3O

+(aq)

MR 1 1 1 1

I 0.22 -- 0 0

C -x --- x x

E 0.22-x -- x x

6

211

2

33

323

102.2

22.0103.2

22.0

][

]][[

x

x

x

x

K

K

NHCH

OHNHCHK

bBase

W

a

[H3O+]=2.2x10-6

pH=-log[H3O+]

pH=5.66

0.22/Ka> 1000

approx will work

0.45/Kb> 1000approx will work


13/20

27.

If 65 mL of 0.20 M calcium hydroxide is need to titrate 85 mL of an unknown concentration ofnitrous acid to equivalence point, calculate:

a) the concentration of the original nitrous acid solution. [0.31 M]b)


2 HNO2 (aq)+ Ca(OH)2 (aq)Ca(NO2)2 (aq)+ 2 H2O (l)c 0.31 0.20 0.087 ([NO2

-]=0.17 M)V 0.085 0.065 0.15n 0.026 0.013 0.013

NO2-aq) + H2O(l) HNO2 (aq) + OH-(aq) Ka =7.2x10-4MR 1 1 1 1

I 0.17 -- 0 0

C -x --- x x

E 0.17-x -- x x

17.0104.1

17.0

][

]][[

211

2

2

2

x

x

x

K

K

NO

OHHNOK

aAcid

W

b

x=1.5x10-6[OH-]=1.5x10-6pOH = -log[OH-]pOH=5.81pH=8.19

Buffer QuestionsFor each of the following questions calculate the pH of a) the initial buffer, b) after the addition of theH3O

+to the initial buffer and c) after the addition of the OH-to the initial buffer.

Question Acid or Base Conjugate Volume (L) moles ofH3O

+moles o

OH-1 0.300 mol of NH3 0.600 mol of NH4NO3 3.0 0.150 0.150

2 0.400 mol HC2H3O2 0.400 mol NaC2H3O2 2.0 0.200 0.100

3 0.250 mol H2CO3 0.200 mol NaHCO3 1.0 0.050 0.060

4 0.200 mol CH3NH2 0.150 mol CH3NH3NO3 3.0 0.020 0.0305 0.900 mol HCN 0.800 mol NaCN 2.0 0.200 0.300

6 1.50 mol benzoic acid 1.20 mol sodium benzoate 3.0 0.600 0.300

7 0.450 mol HClO 0.350 mol KClO 1.0 0.250 0.15

AnswersQuestion Initial pH After H3O

+ After OH-1 8.95 8.56 9.26

2 4.74 4.27 4.96

3 6.26 6.06 6.49

4 10.76 10.65 10.92

5 9.15 8.96 9.476 4.10 3.66 4.30

7 7.43 6.70 7.77

0.17/Kb> 1000

approx will work


14/20

1)

a) NH3 (aq)+ H2O(l)NH4+(aq) + OH

-(aq) Kb=1.8 x 10

-5

MR 1 1 1 1

I 0.10 --- 0.20 0

C -x --- +x +x

E 0.10-x ---- 0.20+x x

6

5

3

4

100.9

)10.0(

)20.0(108.1

][

]][[

x

x

xx

NH

OHNHK

b

[OH-]=9.0x10-6M

pOH = -log[OH-]pOH = 5.05

pH=8.95

b) If you add 0.150 mol of H3O

+ in 3.0 L, the equilibrium will shift right, therefore [NH3] willdecrease to 0.050 M and [NH4

+] will increase to 0.25 M


-(aq) Kb=1.8 x 10

-5

MR 1 1 1 1

I 0.050 --- 0.25 0

C -x --- +x +x

E 0.05-x ---- 0.25+x x

6

5

3

4

106.3

)05.0(

)25.0(108.1

][

]][[

x

x

xx

NH

OHNHK

b


pH=8.56

c) If you add 0.150 mol of OH- in 3.0 L, the equilibrium will shift left, therefore [NH3] will

increase to 0.150 M and [NH4+] will decrease to 0.15 M


-(aq) Kb=1.8 x 10

-5

MR 1 1 1 1

I 0.15 --- 0.15 0

C -x --- +x +x

E 0.15-x ---- 0.15+x x

5

5

3

4

108.1

)15.0(

)15.0(108.1

][

]][[

x

x

xx

NH

OHNHK

b


pH=9.26

0.10/Kb> 1000

approx will work

0.050/Ka> 1000

approx will work

0.15/Ka> 1000

approx will work


15/20

2)

a) HC2H3O2 (aq)+ H2O(l)C2H3O2-(aq) + H3O

+(aq) Ka=1.8 x 10

-5

MR 1 1 1 1

I 0.20 --- 0.20 0

C -x --- +x +x

E 0.20-x ---- 0.20+x x

5

5

232

3232

108.1

)20.0(

)20.0(108.1

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=1.8x10-5

pH=-log[H3O+]

pH=4.74

b) If you add 0.200 mol H3O

+ in 2.0 L, the equilibrium will shift left, therefore [HC2H3O2] willincrease to 0.30 M and [C2H3O2

-] will decrease to 0.10 M


+(aq) Ka=1.8 x 10

-5

MR 1 1 1 1

I 0.30 --- 0.10 0

C -x --- +x +x

E 0.30-x ---- 0.10+x x

5

5

232

3232

104.5

)30.0(

)10.0(108.1

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=5.4x10-5

pH=-log[H3O+]

pH=4.27

c) If you add 0.100 mol of OH- in 2.0 L, the equilibrium will shift right, therefore [HC2H3O2]will decrease to 0.150 M and [C2H3O2

-] will increase to 0.25 M


+(aq) Ka=1.8 x 10

-5

MR 1 1 1 1

I 0.15 --- 0.25 0

C -x --- +x +x

E 0.15-x ---- 0.25+x x

5

5

232

3232

101.1

)15.0(

)25.0(108.1

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=1.1x10-5

pH=-log[H3O+]

pH=4.96

0.20/Ka> 1000

approx will work

0.10/Ka> 1000

approx will work

0.15/Ka> 1000

approx will work


16/20

3)

a) H2CO3 (aq)+ H2O(l) HCO3-(aq) + H3O

+(aq) Ka=4.4 x 10

-7

MR 1 1 1 1

I 0.25 --- 0.20 0

C -x --- +x +x

E 0.25-x ---- 0.20+x x

7

7

32

33

105.5

)25.0(

)20.0(104.4

][

]][[

x

x

xx

COH

OHHCOK

a

[H3O+]=5.5x10-7

pH=-log[H3O+]

pH=6.26

b) If you add 0. 050 mol H3O

+ in 1.0 L, the equilibrium will shift left, therefore [H2CO3] willincrease to 0.30 M and [HCO3


H2CO3 (aq)+ H2O(l) HCO3-(aq) + H3O

+(aq) Ka=4.4 x 10

-7

MR 1 1 1 1

I 0.30 --- 0.15 0

C -x --- +x +x

E 0.30-x ---- 0.15+x x

7

7

32

33

108.8

)30.0(

)15.0(104.4

][

]][[

x

x

xx

COH

OHHCOK

a

[H3O+]=8.8x10-7

pH=-log[H3O+]

pH=6.06

c) If you add 0.060 mol of OH- in 1.0 L, the equilibrium will shift right, therefore [H2CO3] willdecrease to 0.150 M and [HCO3


H2CO3 (aq)+ H2O(l) HCO3-(aq) + H3O

+(aq) Ka=4.4 x 10

-7

MR 1 1 1 1

I 0.19 --- 0.26 0

C -x --- +x +x

E 0.19-x ---- 0.26+x x

7

7

32

33

102.3

)19.0(

)26.0(104.4

][

]][[

x

x

xx

COH

OHHCOK

a

[H3O+]=3.2x10-7

pH=-log[H3O+]

pH=6.49

0.20/Ka> 1000

approx will work

0.15/Ka> 1000

approx will work

0.19/Ka> 1000

approx will work


17/20

4)

a) CH3NH2 (aq)+ H2O(l)CH3NH3+(aq) + OH

-(aq) Kb=4.4 x 10

-4

MR 1 1 1 1

I 0.067 --- 0.05 0

C -x --- +x +x

E 0.067-x ---- 0.05+x x

0109.2050.0

)067.0(

)050.0(104.4

][

]][[

52

4

23

33

xx

x

xx

NHCH

OHNHCHK

b

a=1, b=0.050, c= -2.9x10-5

x1=5.7x10-4 x2=-0.051

[OH-]=5.7x10-4 MpOH = -log[OH-]pOH = 3.24

pH=10.76


+ in 3.0 L, the equilibrium will shift right, therefore [CH3NH2]will decrease to 0.060 M and [CH3NH3

+] will increase to 0.25 M


-(aq) Kb=4.4 x 10

-4

MR 1 1 1 1

I 0.060 --- 0.057 0

C -x --- +x +x

E 0.060-x ---- 0.057+x x

0106.2057.0

)060.0(

)057.0(104.4

][

]][[

52

4

23

33

xx

x

xx

NHCH

OHNHCHK

b

a=1, b=0.057, c= -2.6x10-5

x1=4.5x10-4 x2=-0.057

[OH-]=4.5x10-4 MpOH = -log[OH-]pOH = 3.35

pH=10.65

c) If you add 0.030 mol of OH- in 3.0 L, the equilibrium will shift left, therefore [CH3NH2] willincrease to 0.077 M and [CH3NH3

+] will decrease to 0.040 M


-(aq) Kb=4.4 x 10

-4

MR 1 1 1 1

I 0.077 --- 0.040 0

C -x --- +x +x

E 0.077-x ---- 0.040+x x

0104.3040.0

)077.0(

)040.0(104.4

][

]][[

52

4

23

33

xx

x

xx

NHCH

OHNHCHK

b

a=1, b=0.040, c= -3.4x10-5

x1=8.3x10-4 x2=-0.041

[OH-

]=8.3x10-4

MpOH = -log[OH-]pOH = 3.08

pH=10.92

0.05/Kb< 1000


0.057/Kb< 1000


0.040/ Kb< 1000



18/20

5)

a) HCN(aq)+ H2O(l) CN-(aq) + H3O

+(aq) Ka=6.2 x 10

-10

MR 1 1 1 1

I 0.45 --- 0.40 0

C -x --- +x +x

E 0.45-x ---- 0.40+x x

10

10

3

100.7

)45.0(

)40.0(102.6

][

]][[

x

x

xx

HCN

OHCNK

a

[H3O+]=7.0x10-10

pH=-log[H3O+]

pH=9.15

b) If you add 0.200 mol H3O

+ in 2.0 L, the equilibrium will shift left, therefore [HCN] willincrease to 0.55 M and [CN-] will decrease to 0.30 M

HCN(aq)+ H2O(l) CN-(aq) + H3O

+(aq) Ka=6.2 x 10

-10

MR 1 1 1 1

I 0.55 --- 0.30 0

C -x --- +x +x

E 0.55-x ---- 0.30+x x

9

10

232

3232

101.1

)55.0(

)30.0(102.6

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=1.1x10-9

pH=-log[H3O+]

pH=8.96

c) If you add 0.300 mol of OH- in 2.0 L, the equilibrium will shift right, therefore [HCN] willdecrease to 0.30 M and [CN-] will increase to 0.55 M

HCN(aq)+ H2O(l) CN-(aq) + H3O

+(aq) Ka=6.2 x 10

-10

MR 1 1 1 1

I 0.30 --- 0.55 0

C -x --- +x +x

E 0.30-x ---- 0.55+x x

10

10

232

3232

104.3

)30.0(

)55.0(102.6

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=1.1x10-5

pH=-log[H3O+]

pH=9.47

0.40/Ka> 1000

approx will work

0.30/Ka> 1000

approx will work

0.30/Ka> 1000

approx will work


19/20

6)

a) HC7H5O2 (aq)+ H2O(l)C7H5O2-(aq) + H3O

+(aq) Ka=6.3 x 10

-5

MR 1 1 1 1

I 0.50 --- 0.40 0

C -x --- +x +x

E 0.50-x ---- 0.40+x x

5

5

257

3257

109.7

)50.0(

)40.0(103.6

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=7.9x10-5

pH=-log[H3O+]

pH=4.10


+ in 3.0 L, the equilibrium will shift left, therefore [HC7H5O2]will increase to 0.70 M and [C7H5O2



+(aq) Ka=6.3 x 10

-5

MR 1 1 1 1

I 0.70 --- 0.20 0

C -x --- +x +x

E 0.70-x ---- 0.20+x x

4

5

232

3232

102.2

)70.0(

)20.0(103.6

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=2.2x10-4

pH=-log[H3O+]

pH=3.66

c) If you add 0.300 mol of OH- in 3.0 L, the equilibrium will shift right, therefore [HC2H3O2]will decrease to 0.40 M and [C2H3O2



+(aq) Ka=6.3 x 10

-5

MR 1 1 1 1

I 0.40 --- 0.50 0

C -x --- +x +x

E 0.40-x ---- 0.50+x x

5

5

232

3232

100.5

)40.0(

)50.0(103.6

][

]][[

x

x

xx

OHHC

OHOHCK

a

[H3O+]=1.1x10-5

pH=-log[H3O+]

pH=4.30

0.40/Ka> 1000

approx will work

0.20/Ka> 1000

approx will work

0.40/Ka> 1000

approx will work


20/20

7)

a) HClO(aq)+ H2O(l) ClO-(aq) + H3O

+(aq) Ka=2.9 x 10

-8

MR 1 1 1 1

I 0.45 --- 0.35 0

C -x --- +x +x

E 0.45-x ---- 0.35+x x

8

8

2

3

107.3

)45.0(

)35.0(109.2

][

]][[

x

x

xx

HClO

OHClOK

a

[H3O+]=3.7x10-8

pH=-log[H3O+]

pH=7.43


+ in 1.0 L, the equilibrium will shift left, therefore [HClO] willincrease to 0.70 M and [ClO-] will decrease to 0.10 M

HClO(aq)+ H2O(l) ClO-(aq) + H3O

+(aq) Ka=2.9 x 10

-8

MR 1 1 1 1

I 0.70 --- 0.10 0

C -x --- +x +x

E 0.70-x ---- 0.10+x x

7

8

2

3

100.2

)70.0(

)10.0(109.2

][

]][[

x

x

xx

HClO

OHClOK

a

[H3O+]=2.0x10-7

pH=-log[H3O+]

pH=6.69

c) If you add 0.15 mol of OH- in 1.0 L, the equilibrium will shift right, therefore [HClO] willdecrease to 0.30 M and [ClO-] will increase to 0.50 M

HClO(aq)+ H2O(l) ClO-(aq) + H3O+(aq) Ka=2.9 x 10-8

MR 1 1 1 1

I 0.30 --- 0.50 0

C -x --- +x +x

E 0.30-x ---- 0.50+x x

8

8

2

3

107.1

)30.0(

)50.0(109.2

][

]][[

x

x

xx

HClO

OHClOK

a

[H3O+]=1.7x10-8

pH=-log[H3O+]

pH=7.77

0.35/Ka> 1000

approx will work

0.10/Ka> 1000

approx will work

0.30/Ka> 1000

approx will work

acid base problems solutions

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