Calculating the ionic product of water

Question

In the question I've to find out the ionic product of water using the facts that the degree of dissociation of water at $18\ \mathrm{^\circ C}$ is $1.8\times10^{-9}$.

My attempt

Let the concentration of water be $1\ \mathrm M$.

$$\ce{H2O <=> H+ + OH-}$$

Initial:

$$\begin{align} [\ce{H2O}] &= 1\ \mathrm M\\[6pt] [\ce{H+}] &= 0\ \mathrm M\\[6pt] [\ce{OH-}] &= 0\ \mathrm M \end{align}$$

At equilibrium:

$$\begin{align} [\ce{H2O}] &= (1 - 1.8×10^{-9})\ \mathrm M\\[6pt] [\ce{H+}] &= 1.8×10^{-9}\ \mathrm M\\[6pt] [\ce{OH-}] &= 1.8×10^{-9}\ \mathrm M \end{align}$$

Therefore, $$\begin{align} K_\mathrm w &= \frac{[\ce{H+}] + [\ce{OH-}]}{[\ce{H2O}]}\\[6pt] &= \left(1.8\times10^{-9}\right) \times \left(1.8\times10^{-9}\right)\\[6pt] &= 3.24 \times 10^{-18} \end{align}$$

According to my book answer is $1.0\times10^{-14}$, which I know is correct. I want to know where am I going wrong?

Answer 1

What you did wrong was adding the concentration of the hydron and the hydroxide ion. Also, although it hardly matters, the concentration of $\ce{H_2O}$ is considered a constant and ignored in the product.

Answer 2

What it boils down to is, you need to take the fractional dissociation $1.8×10^{-9}$ and multiply it by the total amount of water, not just one molar. The total concentration of water is $1000$ grams per liter, which you convert to moles per liter by dividing by the molecular weight of the water. After working this out and doing the multiplication you should get the right hydron and hydroxide ion concentrations and then you get the right $K_w$.

This is just a number crunching exercise. To find the ion product for real, we can use electrochemistry. Using a highly reversible cathodic reaction for hydrogen evolution, say on a platinum electrode, find the hydrogen reduction potential for 1 M strong acid and for 1 M strong base, e.g. hydrochloric acud and sodium hydroxide. There is a difference between them, which you plug into the Nernst Equation to get the hydron concentration in the base solution. The ion product then follows.