# Calculating the pH

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)

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?

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 $\mathrm{pH}=6.98$
• @user21398 The first equation is called a charge-balance equation and says that, since the system as a whole must be electrically neutral, the total positive charge in the system must be equal to the total negative charge in the system. The second equation is just replacing $[\ce{Cl-}]$ with the concentration of the strong acid (since it dissociates fully) and replacing $[\ce{OH-}]$ with the expression involving $K_{\text{w}}$. Oct 4 '15 at 15:47