Question

Boric acid is a weak monoprotic acid. It ionizes in water as

$B{\left(OH\right)}_3+H_{2}O\rightleftharpoons B{\left(OH\right)}_4^{-}+H^{+};K_a=5.9\times {10}^{-10}$

Calculate $pH$ of $0.3M$ boric acid.

• A. $4.87$
• B. $6.33$
• C. $9.13$
• D. $10.77$

Answer 1

The correct option is A $4.87$

We know that by Ostwald's Dilution Law,

$K_a={\alpha }^{2}C$

${\alpha }^2=\frac{K_a}{C}=\frac{5.9\times {10}^{-10}}{0.3}=19.6\times {10}^{-10}$

$\alpha =4.434\times {10}^{-5}$

$\left[H^{+}\right]=$$\alpha \times C=$$1\times 4.434\times {10}^{-5}\times 0.3=1.33\times {10}^{-5}mol/d{m}^3$

$pH=-lo{g}_{10}\left[H^{+}\right]$$=-lo{g}_{10}(1.33\times {10}^{-5})$$=$$4.876$