The degree of dissociation of $0.1 \mathrm{M}$ weak acid HA is $0.5 \%$. If $2 \mathrm{~mL}$ of $1.0…
The degree of dissociation of $0.1 \mathrm{M}$ weak acid HA is $0.5 \%$. If $2 \mathrm{~mL}$ of $1.0 \mathrm{MHA}$ solution is diluted to $32 \mathrm{~mL}$ the degree of dissociation of acid and $\mathrm{H}_{3} \mathrm{O}^{+}$ ion concentration in the resulting solution will be respectively
$0.02$ and $3.125 \times 10^{-4}$
$1.25 \times 10^{-3}$ and $0.02$
$0.632$ and $3.95 \times 10^{-4}$
$0.02$ and $8.0 \times 10^{-12}$
Solution
\(a=0.5 \%=0.005 \text { and } C=0.1 \mathrm{M}\)
The expression for the degree of dissociation is \(a=\sqrt{\frac{K_a}{C}}\).
\(\begin{aligned}
& 0.005=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{0.1}} \\
& \mathrm{~K}_{\mathrm{a}}=2.5 \times 10^{-6}
\end{aligned}\)
When the solution is diluted, the molarity of the solution is given by the following expression.
\(M_1 V_1=M_2 V_2\)
or \(2 \times 1=32 \times M_2\)
Hence, \(M_2=\frac{1}{16}\)
For this diluted solution, \(a=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}=0.632 \%\).
The hydrogen ion concentration is \(\left[\mathrm{H}^{+}ight]=\mathrm{ca}=\frac{1}{16} \times 0.00632=3.955 \times 10^{-4}\).
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