Question

The Kw of ${2H}_{2}O\rightleftarrows H_3{O}^{+}+O{H}^{-}$ changes from ${10}^{-14}$ at ${25}^{\circ }C$ to $9.62\times {10}^{-14}$ at ${60}^{\circ }$. What is the pH of water at ${60}^{\circ }C$?

Answer 1

$K_w=\left[H^{+}\right]\left[{OH}^{-}\right]\left[H^{+}\right]\left[{OH}^{-}\right][H^{+}{]}^2=9.62\times {10}^{-14}{H}^{+}=\sqrt{9.62\times {10}^{-14}}=3.1\times {10}^{-7}PH=-log(3.1\times {10}^{-7})=-0.5+7PH=6.5solutionwouldbelittleacidic.$