Question

The pH of a neutral solution at ${50}^{0}C$ is:

$(K_w={10}^{-13.26}at{5}^{0}C)$

• A. 7
• B. 6
• C. 7.23
• D. 6.63

Answer 1

Answer: (D)

6.63

$H_{2}O\left(l\right)\rightleftharpoons H^{+}+O{H}^{-}$

$c\alpha$ $c\alpha$

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

$\left[H^{+}\right]=\left[O{H}^{-}\right]=x$

$\therefore \Rightarrow x^2=K_w$

$\Rightarrow x^2={10}^{-13.26}$

$\Rightarrow x={10}^{-6.63}$

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

$=-log$

${10}^{-6.63}$

$=6.63$