Autoionization of water
Maybe a numerical example will help you. Let's start from pure water, with the following concentrations :$\ce{[H+] = [OH-] = 10^{-7} M}$. Now we will suppose you add $\ce{10^{-7}} mol$ $\ce{HCl}$ in one liter of this water. Suddenly, the concentration of $\ce{H+}$ should double. This is not possible in the long run. A part $a$ of these supplementary $\ce{H+}$ ions must react with the same amount $a$ of the $\ce{OH-}$ ions to form water. The final molar concentrations of both ions are given by  : $$\ce{[H+] = 2.000·10^{-7}} - a$$  $$\ce{[OH-] = 1.000·10^{-7} } - a$$   This amount $a$ may be obtained by solving the equation : $$\ce{ (2.000·10^{-7}} - a)\ce{(1.000 10^{-7}} - a)= 1.000·10^{-14}$$ The solution is : $a = 0.382·10^{-7} mol/L$. So about $38.2$% of the initial $\ce{OH-}$ ions have been destroyed by the addition of the new acidic ions. And the numerical values of the final concentrations are $$\ce{[H+] = 2.00·10^{-7}}- a = 1.618·10^{-7} mol/L$$$$\ce{[OH-] = 1.00 10^{-7} } - a = 0.618· 10^{-7} mol/L$$ And of course : $\ce{[Cl-] = 1.000·10^{-7}$ $ mol/L}$. If you want you may check the product of the two final concentrations $\ce{[H+]}$ and $\ce{[OH-]}$. It is equal to ${1.000·10^{-14} mol^2L^{-2}}$.