At a given temperature, the degree of ionization (fraction dissociated) of water is found to be $$1.8 \times 10^{-9}$$. Calculate the ionization constant and the ionic product of water at this temperature.

PLACEHOLDER ANSWER - the question has two answers. Pure water is $$\frac{1000}{18} \approx 55.5\text{ mol L}^{-1}$$, so $$[\text{H}^+] = [\text{OH}^-] = 55.5 \times 1.8 \times 10^{-9} \approx 1.0 \times 10^{-7}\text{ M}$$. The ionic product is $$K_w = [\text{H}^+][\text{OH}^-] \approx 1.0 \times 10^{-14}$$. The ionization constant is $$K = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]} = \frac{10^{-14}}{55.5} \approx 1.8 \times 10^{-16}$$.