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  4. The pH of 0.001 M Ba(OH)2 solution will be

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Q. The $pH$ of $0.001\,M\, Ba(OH)_2$ solution will be

Equilibrium

Solution:

$Ba(OH)_2 \rightleftharpoons Ba^{2+} +2OH^-$
$\left[OH^{-}\right]=2\times1\times10^{-3}\,M$
$pOH = -log \left[OH^{-}\right] $
$= -log \left(2 \times10^{-3}\right) = 2.7$
$pOH + pH = 14$
$pH = 14-2.7= 11.3$