# Calculating the pH

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

Calculate the pH of $10^{-8}\ \mathrm{mol/L}$ $\ce{HCl}$ solution in water. ($\ce{HCl}$ is a strong acid which completely ionizes in water)

1. completely ionizes in water so: $10^{-8}\,\mathrm{mol}$ $\ce{H+}$ 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: $10^{-8} + 9.51\times10^{-8}$

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

• Okay, thanks for the tips. Still fairly new to this. – user21398 Oct 4 '15 at 13:51
• well is you see the pKw of water at 25 C is 13.99 Wikipedia so they must have used it – Chinmay Chandak Oct 4 '15 at 14:11

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 $K_{\text{w}}$. – orthocresol Oct 4 '15 at 15:47