Question

A weak acid HyA has pk, values of 1.95 (pKsi), 5.08 (pKz), and 10.5 acid was dissolved in deionized water to make 100.0 mL of a 0.015 M solution. (pKsa). The disodium salt of that weak 8. A) What was the pH of the resulting solution? B) What was the equilibrium concentration of H2A? c) what was the value of a?-? CHEM 210, Fall 2017-Problem Set 6 Page 4 of 4

Answer 1

Disodium salt of the acid: Na₂HA

This is an amphoteric salt i.e. it can undergo protonation and deprotonation as well.

HA^2- + H⁺ --> H₂A^-

HA^2- --> A^3- + H⁺

(i)

For this case,

pH = (pKa2 + pKA)/2

= (5.08 + 10.54)/2 = 7.81

(ii)

pH = - log([H⁺])

[H⁺] = 10^-pH

= 10^-7.81 = 1.55 x 10^-8

Ka2 = 10^-5.08 = 8.32 x 10^-6

HA^2- + H⁺ --> H₂A^- ; K = Kw/Ka2

(0.015 – x) + 1.55 x 10^-8 --> x

K = Kw/Ka2 = 10^-14 / 8.32 x 10^-6

K = 1.2 x 10^-9

K = [H₂A^-] / ([H⁺] [HA^2-])

= x / (1.55 x 10^-8 * (0.015 – x)) = 1.2 x 10^-9

Solving we get, x = 2.8 x 10^-19

[H₂A^-] = 2.8 x 10^-19

(iii)

A^3- + H⁺ --> HA^2-

0.015(1 – α) + 1.55 x 10^-8 --> 0.015 α

[HA^2-] = 0.015 - x = 0.015

[HA^2-] = 0.015 α = 0.015

α = 1