1. Tardigrade
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  3. Chemistry
  4. In a buffer solution concentration of (NH4)2SO4 and NH4OH are 0.4 M and 0.8 M respectively, pH of the solution is [ Ka(NH4+) = 10-8]

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Q. In a buffer solution concentration of $(NH_4)_2SO_4$ and $NH_4OH$ are $0.4 \,M$ and $0.8\, M$ respectively, $pH$ of the solution is $[ K_a(NH_{4}^{+}) = 10^{-8}]$

Solution:

$\left(K_{b}\right)_{NH_4OH} =\frac{10^{-14}}{10^{-8}}=10^{-6}$
$\left[NH_{4}^{+}\right]=0.4\times2=0.8\,M$
$\left[NH_{4}OH\right]=0.8\,M$
$pOH=PK_{b}+log\frac{{\text[salt]}}{\text{[base]}}$
$pOH=-log10^{-6}+log\frac{0.8}{0.8}=6$
$PH=14-pOH=14-6=8$