Question

Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found in citrus fruit. Calculate [HAsc–], [Asc2–], and the pH of 0.050 M H2Asc.

Answer 1

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$         $K_{a1}=1.0\times 10^{-5}$

Initial

c                0              0

Equilibrium

c-x                x          x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$        $K_{a2}=5\times 10^{-12}$

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$