Question

a. When a 20.0 mL sample of a 0.431 M aqueous nitrous acid solution is titrated with a 0.369 M aqueous sodium hydroxide solution, what is the pH at the midpoint in the titration?

b. A 26.4 mL sample of a 0.351 M aqueous nitrous acid solution is titrated with a 0.345 M aqueous potassium hydroxide solution. What is the pH at the start of the titration, before any potassium hydroxide has been added?

Answer 1

a)

Answer

pH = 3.40

Explanation

At the mid point of the titration between a weak acid and a strong base , the pH would be equal to the pKa of the weak acid.

Nitrous acid is weak acid and sodium hydroxide is a strong base

Therefore

At the mid point of titration between nitrous acid and sodium hydroxide is equal to pKa of nitous acid

pKa of nitrous acid = 3.398

pH at the point of the titration = 3.40

b)

Answer

pH = 1.93

Explanation

Dissociation equillibrium of HNO2 is

HNO2 (aq) + H2O(l) <-------> NO2^-(aq) + H3O⁺(aq)

K_a = [NO2^-] [ H3O⁺] / [HNO2]

Initial concentration

[HNO2] = 0.351

[NO2^-] = 0

[H3O⁺] = 0

change in concentration

[HNO2] = - x

[NO2^-] = + x

[H3O⁺] = + x

Equilibrium concentration

[HNO2] = 0.351 - x

[ NO2^-] = x

[H3O⁺] = x

pKa = -logKa

-logKa = 3.398^{​​​​​​}

Ka = 4.00 × 10^-4

x²/( 0.351 - x) = 4.00 ×10^-4

solving for x

x = 0.01165

[H3O⁺] = 0.01165

pH = - log [H3O⁺]

pH = - log(0.01165)

pH = 1.93