Calculating the pH

Question

I have a question about a pH calculation which is the following:

Calculate the pH of $\pu{E-8 mol/L}~\ce{HCl}$ solution in water. ($\ce{HCl}$ is a strong acid which completely ionizes in water)

So what I had done:

1. completely ionizes in water so: $\pu{E-8 mol}~\ce{H+}$ is added to water.
2. Water equillibrium says: $\ce{2H2O <=> H3O+ + OH-}$
3. $K_\mathrm w = \ce{[H3O+]} \times \ce{[OH- ]}$
4. therefore I assume $\ce{[H3O+]} = 10^{-7}$, because $(10^{-7}\times 10^{-7} = 10^{-14})$
5. $10^{-7} + 10^{-8} = 1.1\times10^{-7}\,\frac{\mathrm{mol}\,\ce{H3O+}}{\mathrm{L}}$
6. $\mathrm{pH} = -\log(\mathrm{ans}) = 6.96$

However, after I checked the answer it says the $\mathrm{pH}$ is supposed to be $6.98$ instead of $6.96$. What they did was: $10^{-8} + \pu{9.51E-8}$

Where does this $9.51\times10^{-8}$ come from?

Answer 1

We can not in this case neglect water auto-dissociation reaction when determining the pH of the solution.

The equation of the electro-neutrality of the solution: $$\ce{[H3O+]}=\ce{[Cl- ] + [OH- ]}$$ $$\ce{[H3O+]} =c+ \frac{K_\mathrm w}{\ce{[H3O+]}}$$ where $c$ is the concentration of the strong acid.

By arranging the above equation, we get a second order equation: $$\ce{[H3O+]}^2 -c\ce{[H3O+]}- K_\mathrm w=0$$ If we solve this equation (and take only the positive solution), we get: $$\ce{[H3O+]}=\frac{10^{-8}+ \sqrt {10^{-16}+ 4 \times 10^{-14}}}{2}$$ By taking the minus of the decimal logarithm $\mathrm{pH}=6.98$