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  4. The pOH of 0.0005 M sulphuric acid is

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Q. The pOH of $0.0005\, M$ sulphuric acid is

COMEDKCOMEDK 2011Equilibrium

Solution:

$H _2 SO _4 \rightleftharpoons 2 H ^{+}+ SO _4^{2-}$
So, $\left[ H ^{+}\right]=2 \times 0 \cdot 0005=1 \times 10^{-3} M$
$pH =-\log \left[ H ^{+}\right]=-\log \left(1 \times 10^{-3}\right)$
$pH =3$
$\because pH + pOH =14$
$3+ pOH =14$
$ pOH =14-3$
$ pOH =14$
$pOH =11$