Ionic EquilibriumHard
Question
Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 M of NH4Cl and 0.05 M of NH4OH. [Mg2+] in the resulting solution is (Kb for NH4OH = 2.0 × 10–5 and Ksp of Mg(OH)2 = 8.0 × 10–12)
Options
A.4 × 10−6 M
B.2 × 10−6 M
C.0.5 M
D.2.0 M
Solution
$\left\lbrack OH^{-} \right\rbrack = K_{b}.\frac{\left\lbrack NH_{4}OH \right\rbrack}{\left\lbrack NH_{4}^{+} \right\rbrack} = 2 \times 10^{- 5} \times \frac{0.05}{0.25} = 4 \times 10^{- 6}\text{ M}$
$\therefore\left\lbrack Mg^{2 +} \right\rbrack = \frac{K_{sp}}{\left\lbrack OH^{-} \right\rbrack^{2}} = \frac{8 \times 10^{- 12}}{\left( 4 \times 10^{- 6} \right)^{2}} = 0.5\text{ M}$
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