Question

A^2- is the most basic species in a diprotic system. How many grams of HCl must be dissolved in a 1 L solution made from 0.1 mol Na₂A to bring the pH to 5.50?

Ka₁ = 1.00 x 10^-3

Ka₂ = 1.00 x 10^-6

Answer 1

Ka2 = 10^ -6

pka2 = -log Ka2 = -log (10^ -6) = 6

our pH is closer to pka2

we use Henderson equation

pH = pka + log [conjugate base] /[acid]

our conjugate base is A^2- ( Na2A is nothing but A2-) , acid is HA- , pka is pka2

5.5 = 6 + log [A2-] /[HA-]

-0.5 = log [A2-]/[HA-]

[A2-] /[HA] = 10^ -0.5

[A2-] = 0.31623 [HA-]

A2- moles = 0.31623 HA- moles ..............(1)

initially we had 0.1 moles A2-

let HCl moles added be m

then we have reaction H+ (aq) + A2- (aq) <--> HA- (aq)

hence after adding HCl moles     A2- moles = 0.1-m

HA- moles = H+ moles added= m

substituting these in eq (1) we get

0.1-m = 0.31623( m)

m = 0.076

m = moles of H+ = moles of HCl = 0.076

Mass of HCl = moles x molar mass of HCl

        = 0.076 x 36.45 g/mol

         = 2.77 g