At equilibrium \(\left[H^{+}=A^{-}\right]\)
\(K_{a} =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}=\frac{\left[H^{+}\right]^{2}}{[H A]}\)
\(\left[H^{+}\right]=\sqrt{K_{a}[H A]} =\sqrt{1 \times 10^{-5} \times 0.1}\)
\(= \sqrt{1 \times 10^{-6}}=1 \times 10^{-3}\)
\(\alpha =\frac{A \text {ctual ionisation}}{\text {Molar concentration}}\)
\(\%\) of acid dissociated \(=10^{-2} \times 1.00\)
\(=1 \%\)
(આપેલ : $K _{ b }\left( NH _4 OH \right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11=$ $1.04)$
$[$પાણીનો આયનીય ગુણાકાર $ = 1 \times {10^{ - 14}}]$
$\text { Given : pKa }\left( CH _3 COOH \right)=4.76$