AcidsLesson 12

Calculating KaFrom pH

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0CE

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0CE

pH = 1.28

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0CE

pH = 1.28

[H+] = 10-1.28

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0CE

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0CE 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

Ka = [H+][HC2O4-]

[H2C2O4]

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

Ka = [H+][HC2O4-]

= (0.05248)2

[H2C2O4] 0.04752

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the

weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

Ka = [H+][HC2O4-]

= (0.05248)2 = 5.8 x 10-2

[H2C2O4] 0.04752

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0CE

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 0 0CE

pH = 11.427

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0CE

pH = 11.427pOH = 2.573

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0CE

pH = 11.427pOH = 2.573[OH-] = 10-2.573

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0CE

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[OH-] = 0.002673 M

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0CE 0.002673 0.002673

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[H+] = 0.002673 M

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[OH-] = 0.002673 M

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[OH-] = 0.002673 M

Kb = [NH4+][OH-]

[NH3]

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[OH-] = 0.002673 M

Kb = [NH4+][OH-]

= (0.002673)2

[NH3] 0.3973

2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the

Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673

pH = 11.427pOH = 2.573[OH-] = 10-2.573

[OH-] = 0.002673 M

Kb = [NH4+][OH-]

= (0.002673)2 = 1.8 x 10-5

[NH3] 0.3973

3. The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid.

Ka = 7.3 x 10-10

Boric acid

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

C 0.002858 0.002858 0.002858

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

C 0.002858 0.002858 0.002858

E 0.1971 0.002858 0.002858

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

C 0.002858 0.002858 0.002858

E 0.1971 0.002858 0.002858

[OH-] = 0.002858 M

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

C 0.002858 0.002858 0.002858

E 0.1971 0.002858 0.002858

[OH-] = 0.002858 M

Kb = [HCN][OH-]=

[CN-]

4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

CN- + H2O ⇄ HCN + OH-

I 0.20 0 0

C 0.002858 0.002858 0.002858

E 0.1971 0.002858 0.002858

[OH-] = 0.002858 M

Kb = [HCN][OH-]= (0.002858)2 = 4.1 x 10-5

[CN-] 0.1971

5. Calculate the pH of 0.020 M H3BO3

H3BO3 ⇄ H+ + H2BO3-

I 0.020 M 0 0C x x xE 0.020 - x x x

0 small ka

x2 = 3.8 x 10-10

0.020

x = [H+] = 2.76 x 10-6 M

pH = -Log[2.76 x 10-6]

pH = 5.42

2 sig figs due to molarity and Ka

6. Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M HCl with 100 mL of water.

HCl → H+ + Cl-

1(0.050 M) 0.025 M 0.025 M2

pH = 1.60