Question

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

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

Answer 1

Answer: (D)

We know at $25$ degrees $K_w=1\times 10^{-14}$
At $100$ degrees $K_W=55\times 10^{-14}$
$H^{+}=\sqrt{(}55\times 10^{-14})$
$pH=-\log \left[H^{+}\right]$
$pH=-\log \sqrt{(}55\times 10^{-14})$
$\frac{1}{2}\times [-\log 55+14\log 10]$
$\frac{1}{2}\times [-1.74+14]$
$\frac{1}{2}\times 12.26$
$=6.13$