1. Tardigrade
  2. Question
  3. Chemistry
  4. The solubility of Mg ( OH )2 in a particular buffer solution is found to be 0.65 g L -1. Thus, pH of the buffer solution is [Ks p. of .Mg ( OH )2=1.8 × 10-11]

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Q. The solubility of $Mg ( OH )_{2}$ in a particular buffer solution is found to be $0.65\, g\, L ^{-1}$. Thus, $pH$ of the buffer solution is $\left[K_{s p}\right.$ of $\left.Mg ( OH )_{2}=1.8 \times 10^{-11}\right]$

Equilibrium

Solution:

$\left[ Mg ( OH )_{2}\right]=0.65\, gL ^{-1}=\frac{0.65}{58} mol \,L ^{-1}$
$\therefore \left[ Mg ^{2+}\right]=\frac{0.65}{58} M$
$\left[ Mg ^{2+}\right][ OH ]^{2}=K_{s p}=1.8 \times 10^{-11}$
$\therefore \left[ OH ^{-}\right]^{2}=1.8 \times 10^{-11} \times \frac{58}{0.65}$
$=1.606 \times 10^{-9}$
$[ OH^- ]=4.0 \times 10^{-5}$
$pOH =4.4$
$pH =14-4.4=9.6$