Question

In an aqueous solution at ${25}^{o}C$ has twice as many ${OH}^{-}$ as pure water its $pOH$ will be:

• A. $6.699$
• B. $7.307$
• C. $7$
• D. $6.98$

Answer 1

Answer: (A)

$6.699$

Option $A$ is the correct answer.
We know that pure water at 25°C has ${10}^{-7}M$ of $O{H}^{-}$.
According to the problem, the aqueous solution at the same temperature has $2\times {10}^{-7}M$ $O{H}^{-}$

Therefore, its $pOH=-log(2\times {10}^{-7})$

$pOH=6.6989$

$or,pOH=6.699$