Question

Calculate the pH of an aqueous solution of sodium hydrogen phosphite, NaHPO₃.

For phosphorous acid, H₂PO₃, pK_a'= 2.00 and pKa″ = 6.59.

Answer 1

Dissociation of H₂PO₃ in water takes place in two steps.

1. H₂PO₃ (aq) + H₂O (l)  $\rightleftharpoons$ H₃O ⁺ (aq) + HPO₃^- (aq) pKa ' = 2.00

2. HPO₃^- (aq) + H₂O (l)  $\rightleftharpoons$ H₃O ⁺ (aq) + PO₃^2- (aq) pKa " = 6.59

From above reactions, it is clear that HPO₃^- is an intermediate form of H₂PO₃ and PO₃^2-.

We know that, pH of intermediate form of diprotic acid is calculated by using formula pH = 1/2 ( pK ₁ + pK ₂)

$\therefore$ pH of aqueous solution of  NaHPO₃ = 1/2 (  pKa ' + pKa " )

= 1/2 ( 2.00 + 6.59 )

= 1/2 ( 8.59 )

= 4.295

= 4.30

ANSWER : pH of aqueous solution of  NaHPO₃ =4.30