Question

At $90{}^{\circ }C,$ pure water has $H_3{O}^{+}$ ion concentration of ${10}^{-6}mol\text{/}L.$ The $K_w$ at $90{}^{\circ }C$ is

• A. ${10}^{-6}$
• B. ${10}^{-14}$
• C. ${10}^{-12}$
• D. ${10}^{-8}$

Answer 1

Answer: (C)

$K_w=[H^{+}][O{H}^{-}]$ $=({10}^{-6})\times ({10}^{-6})$ $(\because [H^{+}]=[O{H}^{-}])$ $={10}^{-12}$