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# $PH$ of a solution containing 50 mg of sodium hydroxide in $10\;dm^3$ of the solution is

$\begin{array}{1 1} (a)\;9 \\(b)\;3.9031 \\ (c)\;10.0969 \\ (d)\;10 \end{array}$

$P^{OH} =-\log (1.25 \times 10^{-4})$
$\qquad= -0.0969 -4.0=3.9031$
$p^{H} =14-3.9031 =10.0969$