Question

Ascorbic acid is a diprotic acid (Ka1 6.9x10-5) and (ka2 = 2.7x10-12). What is the pH of

a solution that contains 5.0 mg of acid per milliter solution.

Answer 1

Molar mass of ascorbic acid $=176.12g$

Concentration$=\frac {5×10^{-3}}{176.12}×10^3$

$=0.02839M$

$ka_1\ggg\>ka_2$

$\implies$ First dissociation produces virtually all of $H_3O^+$

$H_2C_6H_6O_{6(aq)}+H_2O_{(l)}\rightleftharpoons HC_6H_6O_6^-(aq)+H_3O^+(aq)$

$\begin{matrix} ICE&H_2C_6H_6O_6&HC_6H_6O_6^-&H_3O^+\\ I&0.02839&0&0 \\ C&-x&x&x\\ E&0.02839-x&x&x \end{matrix}$

$6.9×10^{-5}=\frac{(x)(x)}{(0.02839-x)}\>\>\>x\lll0.02839$

$\therefore\>x^2=1.959×10^{-6}$

$x=1.4×10^{-3}$

$PH=-$ log $(1.4×10^{-3})$

$=2.854$