Question

At ${90}^{o}C$ , pure water has $\left[H_3{O}^{+}\right]={10}^{-6}$ m on the value of $k_w$ at this temperature is :

• A. ${10}^{-6}$
• B. ${10}^{-12}$
• C. ${10}^{-13}$
• D. ${10}^{-14}$

Answer 1

Answer: (B)

${10}^{-12}$

Given :- concentration of $H_3{O}^{+}={10}^{-6}mol/L$

Ionization reaction of $H_{2}O$ is :- $H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

$K=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}\Rightarrow K\times \left[H_{2}O\right]=\left[H^{+}\right]\left[O{H}^{-}\right]$

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

$\left[H^{+}\right]=\left[O{H}^{-}\right]={10}^{-6}mol/6$ ($\therefore$ it mentioned that it is pure water)

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-6}\times {10}^{-6}={10}^{-12}$

$\therefore$ Ans is B) ${10}^{-12}$