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 $K _{ W }=55 \times 10^{-14}$
$\left.H ^{+}=\sqrt{(} 55 \times 10^{-14}\right)$
$pH =-\log \left[ H ^{+}\right]$
$\left.pH =-\log \sqrt{(} 55 \times 10^{-14}\right)$
$\frac{1}{2} \times[-\log 55+14 \log 10]$
$\frac{1}{2} \times[-1.74+14]$
$\frac{1}{2} \times 12.26$
$=6.13$