Question

If an aqueous solution at ${25}^{\circ }$C has twice as many $O{H}^{-}$ as pure water its pOH will be

• A. 6.699
• B. 7.037
• C. 7
• D. 6.98

Answer 1

The correct option is A 6.699

$\left[O{H}^{-}\right]=2\times {10}^{-7}$
$log\left[O{H}^{-}\right]=-7+log2$
$-log\left[O{H}^{-}\right]=7–log2=6.699$