Question

The pH of the solution at $25$ which has twice as many hydroxide ion as in pure water at $25$, will be:

• A. 14
• B. 9
• C. 6.699
• D. 7.301

Answer 1

Answer: (D)

7.301

Given,

$Temp={25}^{0}C$

So, No of hydroxide in pure water is ${10}^{-7}$

A/C to this condition no of hydroxide ion is $2\times {10}^{-7}$

So, $poH=-lo{g}_{10}[2\times {10}^{-7}]$

$7+0.301=7.301$