Ionic EquilibriumMedium
Question
pH of 0.01 M-(NH4)2SO4 and 0.02 M-NH4OH buffer (pKa of NH4+ = 9.26) is
Options
A.9.26 + log 2
B.9.26 – log 2
C.4.74 + log 2
D.9.26
Solution
$P^{H} = P^{K_{a}\left( NH_{4}^{+} \right)} + \log\frac{\left\lbrack NH_{3} \right\rbrack_{0}}{\left\lbrack NH_{4}^{+} \right\rbrack_{0}} = 9.2 + \log\frac{0.02}{0.01 \times 2} = 9.26$
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