Question

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

• A. 6.69
• B. 7.3
• C. 7
• D. 6.98

Answer 1

Answer: (A)

6.69

$\left[O{H}^{-}\right]$ for pure water at ${25}^{o}C={10}^{-7}$

$\left[O{H}^{-}\right]$ for the aqueous at ${25}^{o}C=2\times {10}^{-7}$

Hence, $pOH-log\left[O{H}^{-}\right]=-log(2\times {10}^{-7})=7-log2=6.698$