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 $1\ \mathrm M$.

$$\ce{H2O &lt;=&gt; H+ + OH-}$$

Initial:

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

At equilibrium:

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

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

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