Question

A weak monobasic acid is 1% ionized in 0.1 M solution at $25{}^{\circ }C$ . The percentage of ionisation in its 0.025 M solution is

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (B)

lionization of weak acid at $C_1=0.1M$ $\begin{array}{cccc}HA& H^{+}& (aq)+& A^{-}(aq)\\ C_1& 0& 0& :Initialconcentration\\ C_1(1-{\alpha }_1)& C_1{\alpha }_1& C_1{\alpha }_1& :Conc.atequilibrium\end{array}$ $\therefore$ $K_a=\frac{C_1{\alpha }_1.C_1{\alpha }_1}{C_1(1-{\alpha }_2)}=\frac{C_1{\alpha }_1}{(1-{\alpha }_2)}$ $=C_1{\alpha }_1^2$ $(\because {\alpha }_2<<<1)$ lionization of weak acid $C_2=0.025M$ $\begin{array}{cccc}HA& H^{+}& (aq)+& A^{-}(aq)\\ C_2& 0& 0& :Initialconcentration\\ C_2(1-{\alpha }_2)& C_2{\alpha }_2& C_2{\alpha }_2& :Conc.atequilibrium\end{array}$ $\therefore$ $K_{\alpha }=\frac{C_2{\alpha }_2.C_2{\alpha }_2}{C_2(1-{\alpha }_2)}=\frac{C_2{\alpha }_2^2}{(1-{\alpha }_2)}$ $=C_2{\alpha }_2^2$ $(\because {\alpha }_2<<<1)$ $\because$ lionization constant of an acid is a constant and does not change with concentration. $\therefore$ $C_2{\alpha }_1^2=C_2{\alpha }_2^2$ Or ${\alpha }_2^2=\frac{C_1{\alpha }_1^2}{C_2}=\frac{0.1\times {(1)}^2}{0.025}$ Or ${\alpha }_2^2=4$ Or ${\alpha }_2=2$ $\therefore$ The percentage of ionization of weak acid in 0.025 M solution is 2%.