Question

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

• (A) ${10}^{-12}$
• (B) ${10}^{-6}$
• (C) ${10}^{12}$
• (D) ${10}^6$

Answer 1

The Correct Option is A

Explanation:
Given: Concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}={10}^{-6}\mathrm{mol}/\mathrm{L}$Ionization reaction of ${\mathrm{H}}_2\mathrm{O}$ is : $-{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$$K=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}$$\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}\mathrm{mol}/l(\because$ it mentioned that it is purewater )$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-6}\times {10}^{-6}={10}^{-12}$Hence, the correct option is (A).