Question

Ionic product of water at 310 k is $2.7\times {10}^{-14}$ What is the neutral pH of water at this temperature?

• A. 6.9
• B. 6.8
• C. 8.6
• D. 6.6

Answer 1

The correct option is B 6.8

Ionic product of water $=2.7\times {10}^{-14}$
$H_{2}O\to H^{+}+O{H}^{-}$
x x
$\left[H^{+}\right]\left[O{H}^{-}\right]=2.7\times {10}^{-14}$
$x^2=2.7\times {10}^{-14}$
$x=\sqrt{2.7\times {10}^{-14}}=1.7\times {10}^{-7}$
$pH-log\left[H^{+}\right]$
$=-log[1.7\times {10}^{-7}]=[0.2304-7]$
$=+6.7696$
$pH=6.8$