Question

Boric acid, $B(OH{)}_3$ is a monobasic Lewis acid. It gives $H^{+}$ ion in its aqueous solution as follows
$B(OH{)}_3+H_{2}O\rightleftharpoons [B(OH{)}_4{]}^{-}+H^{+},K_a=5.9\times {10}^{-10}$. What will be the $P^H$ of its $0.025M$ aqueous solution?

• A. $5.42$
• B. $4.52$
• C. $2.45$
• D. $4.675$

Answer 1

The correct option is A $5.42$

$B{\left(OH\right)}_3+H_{2}O\rightleftharpoons \left[B{\left(OH\right)}_4^{-}\right]+H^{+}0.025M0.025(1-\alpha )0.025\alpha 0.025\alpha K_a=\frac{\left[H^{+}\right]\left[B{\left(OH\right)}_4^{-}\right]}{\left[B{\left(OH\right)}_3\right]}=\frac{\left[0.025\alpha \right]\left[0.025\alpha \right]}{\left[0.025\right(1-\alpha \left)\right]}$

As $\alpha <<1$, so $1-\alpha \simeq 1$

$⟹K_a=0.025{\alpha }^2⟹5.9\times {10}^{-10}=0.025{\alpha }^2⟹\alpha =1.53\times {10}^{-4}\left[H^{+}\right]=0.025\alpha =3.84\times {10}^{-6}pH=-\log \left[H^{+}\right]=-\log [3.84\times {10}^{-6}]=5.42$