# Calculating the ionic product of water

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?

• why concentration of water be assumed 1M?
– JM97
Aug 26, 2016 at 15:22
• @JM97 They've asked to find the ionic product, so I thought that assuming the initial concentration of water to be 1M will simplify the problem. But I've done something wrong I guess. I don't know where am I going wrong. Aug 26, 2016 at 15:32
• molarity is number of moles per litre so please consider the number of moles of water present per liter . Hint: 1kg of water occupies one litre space.
– JM97
Aug 26, 2016 at 15:34
• @JM97 There is no such thing as the "number of moles"; it is called amount of substance, just like mass is not called "number of kilograms". Apr 5, 2018 at 11:34

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.
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 $K_w$.