Question

calculate ph of 0.200 mol/l of H3PO4 (aq) if ka = 2.00x10-2

Answer 1

The reaction is:-

$H_3PO_4+H_2O\rightarrow H_3O^++H_2PO_4^-$

$\text{at t }=0$ $0.2$ 0 0

Let at time t' The concentration of $H_3O^+$ became x

$\text{ at t }=t'$ $0.2-x$ $x$ $x$

So, $k_a=\dfrac{[H_3O^+][H_2PO_4^-]}{[H_3PO_4]}$

$2\times 10^{-2}=\dfrac{x.x}{0.2-x}$

Solving for $x:$

$0.02(0.2-x)=x^2$

$\Rightarrow x^2+0.02x-0.004=0$

From above we get $x=0.01$

$[H_3O^+]=0.01$

$P_H=-log[H_3O^+]$

$=-log(0.01) \\=2$

Hence $P_H$ is 2.