Question

pH of pure water at ${37}^{o}C$ is
(assume, $K_w=2.0\times {10}^{-14};log2=0.30)$

• A. 6.55
• B. 7
• C. 7.25
• D. 6.85

Answer 1

Answer: (A)

6.55

According to question,

$2\times {10}^{-14}=\left[H^{+}\right]\left[O{H}^{-}\right]$

$2\times {10}^{-14}=[H^{+}{]}^2$

$\left[H^{+}\right]=\sqrt{(}2)\times {10}^{-7}$

$\left[H^{+}\right]=1.4\times {10}^{-7}$

$pH=-log\left(H^{+}\right)$

$\Rightarrow -log(1.4\times {10}^{-7})$

$\therefore pH=6.85$

Hence, option D is correct.