Question

pH of $0.01 M (NH_4)_2 SO_4$ and $0.02 \,M\, NH_4OH $ buffer $(pK_a of NH^+_4 = 9.26)$ is

• A. 4.74 + log 2
• B. 4.74 - log 2
• C. 9.26 +log 2
• D. 9.26 + log 1

Answer 1

Answer: (D)

$pK_{a}\left(NH_{4}^{+}\right)=9.26$
$pK_{b}\left(NH_{3}\right)=14-9.26=4.74$
$\left[NH_{4}^{+}\right]=2\times\left[\left(NH_{4}\right)_{2}SO_{4}\right]$
$=2\times0.01=0.02\,M $
$pOH=pK_{b}+log \frac{\left[NH_{4}^{+}\right]}{\left[NH_{4}OH\right]}$
$pOH=4.74+log \frac{0.02}{0.02}=4.74-log\,1$
$pH=14-pOH=14-4.74+log\,1$
$=9.26+log\,1$