Question

The value of $K_w$ at the physiological temperature(${37}^{o}C$) is $2.56\times {10}^{-14}$. What is the $pH$ at the neutral point of water at this temperature?

$(log2=0.3)$

Answer 1

Given $K_w=2.56\times {10}^{-14}$

$K_w={\left[H^{+}\right]}^2\Rightarrow 2.56\times {10}^{-14}={\left[H^{+}\right]}^2$

$\left[H^{+}\right]=1.6\times {10}^{-7}$

$pH=-log\left[H^{+}\right]$

$=-log[1.6\times {10}^{-7}]$

$=7-0.20$

$pH=6.8$