Question

A weak acid $H_{3}A$ has the three consecutive dissociation constants $K_1,K_2$ and $K_3$ in water as $1.0\times {10}^{-4},1.0\times {10}^{-8}$ and $1.0\times {10}^{-16}$ respectively at room temperature. A metal $M$ forms salt $MA$ which is sparingly soluble (solubility product $\sim {10}^{-30})$ in water. It also forms a water soluble chloride $MC{l}_3$ which dissociates into $M^{3+}$ and $C{l}^{-}$ ions completely in aqueous solution. For an aqueous solution of the acid $H_{3}A$ of concentration $C(\sim 1M)$ at room temperature, the correct statement(s) is/ are__________.

• A. The acid acts as a dibasic acid
• B. The $H^{+}$ ion concentration in the solution is close to $(K_{1}C{)}^{1/2}$
• C. The $H^{+}$ ion concentration in the solution is $(3K_1{K}_2{K}_{3}C{)}^{1/4}$
• D. On addition of excess of the soluble salt $MC{l}_3$ to the acid solution, the $H^{+}$ ions concentration in the solution will be $3C$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The acid acts as a dibasic acid
B The $H^{+}$ ion concentration in the solution is close to $(K_{1}C{)}^{1/2}$
D On addition of excess of the soluble salt $MC{l}_3$ to the acid solution, the $H^{+}$ ions concentration in the solution will be $3C$

Given that,

$K_1={10}^{-4},K_2={10}^{-8},K_3={10}^{-16}$

Weak acid $=H_{3}A\sim 1M$

Dissociation of weak acid $H_{3}A$

$H_{3}A\to H^{+}+H_2{A}^{-}$

$K_1=\frac{\left[H^{+}\right]\left[H_2{A}^{-}\right]}{\left[H_{3}A\right]}$ $[\because H^{+}=H_2{A}^{-}]$

$[H^{+}{]}^2=K_1[H_{3}A]=K_{1}C$

$H^{+}=(K_{1}C{)}^{\frac{1}{2}}$

$H_{3}A$ is tribasic acid

$H_{3}A⟶3H^{+}+A^{3-}$

When $MC{l}_3$ added to $H_{3}A$

$\underset{C}{H_{3}A}+MC{l}_3⟶\underset{3C}{3HCl}+MA$

$H^{+}=3C$

option - B,D are correct