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6-2.Equilibrium-II (Ionic Equilibrium)
medium
A buffer solution contains $1\ mole$ of ammonium sulphate and $1\ mole$ of $NH_4OH$ ($K_b = 10^{-5}$). The $pH$ of solution will be
A
$5$
B
$9$
C
$5.3$
D
$8.7$
Solution
$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Cation}]}{[\mathrm{Base}]}$
$\left[\mathrm{NH}_{4}^{+}\right]=2 \times$ mole of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$
$\therefore \mathrm{pOH}=5+\log 2=5.3$
or $\mathrm{pH}=8.7$
Standard 11
Chemistry
