acid and base equilibria

Question

1) Calculate the pH of the resulting solution when 100.0 mL of 0.2 M HNO₂ is titrated by 0.1 M NaOH after

a) 0 mL of NaOH has been added.

b) 100 mL of NaOH have been added.

c) 150 mL of NaOH have been added.

d) 200 mL of NaOH have been added. 

e) 300 mL of NaOH have been added.

 The K_a of HNO₂ is 4.5 x 10^-4

Answer 1

If you are wondering how @Lisandro H arrived at those answers, here is an abbreviated explanation. This problem has several parts, and each takes time to answer. So, here is the short version.

a). 0 NaOH added. pH is just -log [H+] of 0.2 M HNO2

HNO2 ==> H+ + NO2-

Ka = [H+][NO2-] / [HNO2] = (x)(x) / 0.2-x (assume x is small relative to 0.2 M and ignore it)

4.5x10^-4 = x2 / 0.2

x = [H+] = 9.49x10^-3 M

pH = -log 9.49x10^-3

pH = 2.02

b). after addition of 100 ml 0.1 M NaOH

moles HNO2 initially present = 100 ml x 1 L / 1000 ml x 0.2 mol/L = 0.02 moles

moles NaOH added = 100 ml x 1 L / 1000 ml x 0.1 mol/L = 0.01 moles

This is 1/2 the equivalence point and forms a buffer

HNO2 + NaOH ==> NaNO2 + H2O

0.02......0.01...............0...................Initial

-0.01....-0.01...........+0.01...............Change

0.01........0.................0.01..............Equilibrium

Final concentrations:

[HNO2] = 0.01 mol / 200 ml x 1000 ml / L = 0.05 M

[NO2-] = 0.01 mol / 200 ml x 1000 ml / L = 0.05 M

pH = pKa (see Henderson Hasselbalch equation)

pH = 3.35

(c). after addition of 150 ml 0.10 M NaOH

HNO2 + NaOH ==> NaNO2 + H2O

0.02.....0.015...............0.................Initial

-0.015...-0.015........+0.015...........Change

0005.......0................0.015...........Equilibrium

Final concentrations:

[HNO2] = 0.005 mol / 250 ml x 1000 ml/L = 0.02 M

[NO2-] = 0.015 mol / 250 ml x 1000 ml/L = 0.06 M

Henderson Hasselbalch equation: pH = pKa + log [NO2-]/[HNO2]

pH = 3.35 + log (0.06/0.02)

pH = 3.83

(d). after addition 200 ml 0.10 M NaOH

HNO2 + NaOH ==> NaNO2 + H2O

0.02......0.02..............0.................Initial

-0.02...-0.02............+0.02............Change

0.............0...............0.02.............Equilibrium

Now you have 0.02 mols NaNO2 in a total volume of 300 mls (0.300 L) = 0.0667 M NaNO2

Hydrolysis of NO2-: NO2- + H2O ==> HNO2 + OH-

pKb for NO2- = 1x10^-14 / 4.5x10^-4 = 2.22x10^-11

2.22x10^-11 = [HNO2][OH-] / [NO2-]

2.22x10^-11 = (x)(x) / 0.0667-x (assume x is small and ignore it)

2.22x10^-11 = x2/0.0667

x = [OH-] = 1.22x10^-6 M

pOH = 5.91

pH = 8.09

e). after addition of 300 ml 0.10 M NaOH

HNO2 + NaOH ==> NaNO2 + H2O

0.02.......0.03...............0.................Initial

-0.02....-0.02.............+0.02............Change

0...........0.01..............0.02.............Equilibrium

Now you have 0.01 moles NaOH in a volume of 400 mls (0.4 L) = 0.025 M NaOH

[OH-] = 0.025 M

pOH = 1.60

pH = 12.4

Answer 2

The pH of the resulting solution is:

a) 2.0332

b) 3.3545

c) 3.8282

d) 8.0868

e) 12.398

Each item requires a different calculation [except for b) and c) which are done in the same way].