Question

The hydrolytic constant $k_h$ for the hydrolytic equilibrium $H_{2}P{O}_4^{-}+H_{2}O\to H_{3}P{O}_4+O{H}^{-}$ is $1.4\times {10}^{-12}$ , the value of ionization constant for the reaction $H_{3}P{O}_4+H_{2}O\to H_{2}P{O}_4^{-}{H}_3{O}^{+}$ is :

• A. $7.14\times {10}^{-3}$
• B. $7.14\times {10}^{-2}$
• C. $7.14\times {10}^{-1}$
• D. $7.14\times {10}^{-5}$

Answer 1

Answer: (A)

$7.14\times {10}^{-3}$

The relationship betwen the ionization constant $K_a$, the hydrolysis constant $K_h$ and the ionic product of water $K_w$ is $K_a=\frac{K_w}{K_h}$.
Substitute values in the above expression.
$K_a=\frac{1\times {10}^{-14}}{1.4\times {10}^{-12}}=7.14\times {10}^{-3}$
Hence, the value of the ionization constant is $7.14\times {10}^{-3}$.