A) 1
B) 2
C) 3
D) 4
E) 5
Correct Answer: B
Solution :
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%.
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