(Solved): Pure water undergoes self-ionisation according to the equation H2O(I)H(aq)+OH(aq) The e ...

Pure water undergoes self-ionisation according to the equation $${\mathrm{H}}_2\mathrm{O}(\mathrm{I})?{\mathrm{H}}^{?}(\mathrm{aq})+{\mathrm{OH}}^{?}(\mathrm{aq})$$ The equilibrium constant for the reaction as written is $$1.0\times 10^{?14}$$ at $$25^{?}\mathrm{C}$$ and $$1.0\times 10^{?13}$$ at $$60^{?}\mathrm{C}$$. 56 At $$60^{?}\mathrm{C}$$, pure water has a $$\mathrm{pH}$$ value that is A less than $$7.0$$ but the water is still neutral. B less than $$7.0$$ and therefore the water is acidic. equal to $$7.0$$ and therefore the water is neutral. D greater than $$7.0$$ and therefore the water is basic.