Question

Calculate the $pH$ (nearest integer) of a solution containing $6.18g$ $H_3{BO}_3$ and $15.08g$ $Na{H}_2{BO}_3$ in one litre. $K_{H_3{BO}_3}=7.3\times {10}^{-10}$.

Answer 1

First, calculate the pKa of $H_{3}B{O}_3$.

$pKa=-logKa=-log7.3\times {10}^{-10}=9.1367$.

Now calculate the pH of the acidic buffer solution.

$pH=pKa+log\frac{\left[Na{H}_{2}B{O}_3\right]}{\left[H_{3}B{O}_3\right]}=9.1367+log\left(\frac{0.1795}{0.1}\right)=9.1367-0.254=8.88\approx 9$

Note: $n_{\left[H_{3}B{O}_3\right]}=\frac{mass}{molarmass}=\frac{6.18}{62}\approx 0.1mol$. Similary, $n_{\left[Na{H}_{2}B{O}_3\right]}=0.1795mol$