Question

At ${100}^{0}C$ the $K_M$ of water is $55$ times its value at ${25}^{0}C$. What will be the $pH$ of neutral solution?
$(log55=1.74)$

• A. $6.13$
• B. $7.00$
• C. $7.87$
• D. $5.13$

Answer 1

The correct option is A $6.13$

We know at 25 degrees $K_w=1\times {10}^{-14}$

At 100 degrees $K_w=55\times {10}^{-14}$

$H^{+}=\sqrt{(}55\times {10}^{-14})$

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

$pH=-log\sqrt{(}55\times {10}^{-14})$

$\frac{1}{2}\times [-log55+14log10]$

$\frac{1}{2}\times [-1.74+14]$

$\frac{1}{2}\times 12.26$

$=6.13$