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

Answer 1

Answer: (B)

2

Answer 2

lionization of weak acid at ${{C}_{1}}=0.1\text{ }M$ $\begin{matrix} HA & {{H}^{+}} & (aq)+ & {{A}^{-}}(aq) \\\ {{C}_{1}} & 0 & 0 & :Initial\,concentration \\\ {{C}_{1}}(1-{{\alpha }_{1}}) & {{C}_{1}}{{\alpha }_{1}} & {{C}_{1}}{{\alpha }_{1}} & :Conc.\,at\,equilibrium \\\ \end{matrix}$ $\therefore$ ${{K}_{a}}=\frac{{{C}_{1}}{{\alpha }_{1}}.{{C}_{1}}{{\alpha }_{1}}}{{{C}_{1}}(1-{{\alpha }_{2}})}=\frac{{{C}_{1}}{{\alpha }_{1}}}{(1-{{\alpha }_{2}})}$ $={{C}_{1}}\alpha _{1}^{2}$ $(\because {{\alpha }_{2}}<<<1)$ lionization of weak acid ${{C}_{2}}=0.025\text{ }M$ $\begin{matrix} HA & {{H}^{+}} & (aq)+ & {{A}^{-}}(aq) \\\ {{C}_{2}} & 0 & 0 & :Initial\,concentration \\\ {{C}_{2}}(1-{{\alpha }_{2}}) & {{C}_{2}}{{\alpha }_{2}} & {{C}_{2}}{{\alpha }_{2}} & :Conc.\,at\,equilibrium \\\ \end{matrix}$ $\therefore$ ${{K}_{\alpha }}=\frac{{{C}_{2}}{{\alpha }_{2}}.{{C}_{2}}{{\alpha }_{2}}}{{{C}_{2}}(1-{{\alpha }_{2}})}=\frac{{{C}_{2}}\alpha _{2}^{2}}{(1-{{\alpha }_{2}})}$ $={{C}_{2}}\alpha _{2}^{2}$ $(\because {{\alpha }_{2}}<<<1)$ $\because$ 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%.