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}}=0.1\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%.