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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Dissociation constant for a monobasic acid is $10^{-4}$ . What is the $pH$ of the monobasic acid ? (If $\%$ dissociation $= 2\,\%$ )
A
$3.2$
B
$2$
C
$2.3$
D
$5$
Solution
$\mathrm{K}_{\mathrm{a}}=10^{-4}=\mathrm{C}_{0} \alpha^{2} \quad \% \alpha=2 \%$
$\alpha=\frac{2}{100}=0.02$
$\because C_{0}=\frac{K_{a}}{\alpha^{2}}=\frac{10^{-4}}{0.02 \times 0.02}=\frac{1}{4}$
$\left[\mathrm{H}^{+}\right]=\mathrm{C}_{\mathrm{o}} \alpha$
$=\frac{1}{4} \times 0.02=0.005 \Rightarrow 5 \times 10^{-3}$
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$
$=-\log \left[5 \times 10^{-3}\right]$
$=-[\log 5-3 \log 10]$
$=2.3$
Standard 11
Chemistry
