# Calculating PH of solution after adding strong acid (equilibrium)

Consider the classic example when talking about acids and bases:

Say you have 1M of $\ce{HCL}$ and you throw it into water, calculate the resulting pH (= concentration of $\ce{H3O+}$).

The way this problem is done: just take $-log(1) = 0 \rightarrow$ PH of resulting solution is $0$.

I don't understand why this is a valid approach. If you throw a concentration of a strong acid in water, wouldn't $\ce{H2O}$ first establish a new equilibrium with (the new total of) $\ce{H3O+}$ and $\ce{HO-}$ according to the autoionization reaction of water?

$$Kw=[\ce{H3O+}][\ce{HO−}]$$

It seems to me like we're mixing up the reaction quotiënt and the equilibrium constant. If this is right, why can we do this? I know that the autoionization doesn't happen very much but we still use the expression for $Kw$ (wich stays constant) to calculate the concentration of $\ce{HO-}$.

I would think the way to solve this problem is to wait till the new equilibrium has been reached (new concentration of $\ce{H3O+}$) and then we can calculate the pH of the solution.

If the reaction quotiënt and equilibrium constant are negligibly close to eachother, could someone show the proces of throwing the acid in the water in the style of the second answer to the following question please? How can the equilibrium shift, while Kc remains constant?

In order for you to completely understand this, you first have to write the mass balance, charge balance and equilibrium constants. This manner of solving $pH$ always works. Mass balance: $$[HCl]_0=[HCl]+[Cl^-]=[Cl^-]$$ Since $HCl$ is a strong acid, there are no free molecules of $HCl$ after equilibrium has been reached, just chloride ions and hydronium ions. Here, the index $0$ means before equilibrium has been reached.
Charge balance: $$[H_3O^+]=[OH^-]+[Cl^-]$$
Equilibrium constants: $$K_w=[H_3O^+][OH^-]$$ $$K_a=[H_3O^+][Cl^-]\rightarrow\infty$$ I.e., as $HCl$ is a strong acid, it's acid constant tends to infinity for our purposes. Solving this system of equations we get: $$1=[Cl^-]$$ $$[H_3O^+]=\frac{10^{-14}}{[H_3O^+]}+1$$ $$[H_3O^+]^2-[H_3O^+]-10^{-14}=0$$ You might want to solve this second degree equation, find the concentration of hydronium ions and find the exact $pH$. But you will notice that it is very close to $0$. That is because the term $\frac{10^{-14}}{[H_3O^+]}$ in the last equation, which means the concentration of hydroxyl ions on equilibrium is very small compared to the others. So you can approximate it to $0$ when comparing it to the others and get: $$[H_3O^+]=1$$