Question

Calculate $K_a$ for a dibasic acid if its concentration is $0.05N$ and hydrogen ion concentration is $1\times {10}^{-3}mol{L}^{-1}$.

Answer 1

$H_{2}A\rightleftharpoons 2H^{+}+A^{2-}$

$C$

$C-C\alpha$ $2C\alpha$ $C\alpha$

$C=\frac{0.05}{2}=0.025M$

$K_a=\frac{{\left[H^{+}\right]}^2\left[A^{-}\right]}{H_{2}A}=\frac{{\left(2C\alpha \right)}^{2}C\alpha }{C-C\alpha }\approx \frac{4C^2{\alpha }^3}{1}$

$\left[H^{+}\right]=2C\alpha =0.05\alpha ={10}^{-3}$

$\alpha =0.02$

So $K_a=\frac{4C^2{\alpha }^3}{1}=4{\left(0.025\right)}^2{\left(0.02\right)}^3=2\times {10}^{-8}$