MCQ
What is the molar solubility of $Mn(OH)_2$ $(K_{sp} = 4.5\times10^{-14})$ in a buffer solution contain equal moles of $NH_4^+$ and $NH_3$ $(K_b = 1.8\times10^{-5})$ ?
  • A
    $3.0\times10^{-4}$
  • $1.38\times10^{-4}$
  • C
    $1.3\times10^{-3}$
  • D
    $7.3\times10^{-4}$

Answer

Correct option: B.
$1.38\times10^{-4}$
b
Basic buffer $\mathrm{pOH}=\mathrm{pkb}+\log \frac{\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]}$

$=4.74$

$\left[\mathrm{OH}^{-}\right]=1.80 \times 10^{-5}$

$\mathrm{K}_{\mathrm{sp}}, \mathrm{Mn}(\mathrm{OH})_{2}=\mathrm{S} \times[\mathrm{OH}]^{2}$

$\therefore S=\frac{k_{s p}}{\left[O H^{-}\right]^{2}}=\frac{4.5 \times 10^{-14}}{1.8^{2} \times 10^{-10}}$

$S=1.38 \times 10^{-4}$

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