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Q. A buffer solution is prepared in which the concentration of $ NH_3 $ is $0.30 \,M$ and the concentration of $ NH_4^+ $ is $0.20 \,M$. If the equilibrium constant, $ K_b $ for $ NH_3 $ equals $ 1.8 \times 10^{-5} $ , what is the $pH $ of this solution? $ (log \,2.7 = 0.43 ) $

AIPMTAIPMT 2011Equilibrium

Solution:

$pOH = pK _{b}+\log \frac{[\text { salt] }}{\text { [base] }}$
$=-\log K_{b}+\log \frac{[\text { salt] }}{\text { [base] }}$
$=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}$
$=5-0.25+(-0.176)$
$=4.75-0.176=4.57$
$\therefore pH =14-4.57=9.43$