The HCl in an aqueous solution will dissociate according to this equation:
HCl (aq) -----> H⁺ (aq) + Cl⁻ (aq)
The equilibrium constant of that reaction is given, it's value is Ka = 1.3 x 10⁶. We can represents the concentration of the products divided by the concentration of the acid at the equilibrium.
Ka = [H⁺] * [Cl⁻]/[HCl]
Then if we take a look at that value, it's a huge value (ka = 1300000). What does that mean? That the concentration of the products is much more greater than the concentration of the reactant. So we can consider that's completely dissociated. That's why HCl is a strong acid.
The pH of a strong acid is calculated using this formula:
pH = -log [H⁺]
The HF in an aqueous solution will dissociate according to this equation:
HF (aq) <-----> H⁺ (aq) + F⁻ (aq)
The equilibrium constant of this reaction is Ka = 6.6 x 10⁻⁴, and it can be represented by this formula:
Ka = [H⁺] * [F⁻]/[HF]
And since we can't do the assumption that we did before (since the Ka is not a huge number), we have to apply the ICE table to find the pH of this weak acid.
Let's suppose that the initial concentration of our acid is A. And we will represent the change in concentration using the letter X.
HF (aq) <-----> H⁺ (aq) + F⁻ (aq)
I: A 0 0
C: -X +X + X
E: A - X +X + X
How do we read this table? I stands for initial. So, we drop some grams of HF into some water, and the initial concentration that we have of HF (before it is dissociated) is A. Since it is not dissociated the concentration of the products (H+ and F-) are zero.
Then the HF will start dissociating, and the C is the change. Some HF will dissociate into H+ and F- ions. How much? We don't know, that's why we usually use the x letter. But an amount of HF will dissociate (-X, the negative is because it disappears) into an amount of its ions (+X)
Finally what we need are the equilibrium concentrations. The equilibrium concentrations are in the E row (equilibrium). We intiailly had a concentration of HF that was A, and a concentration that is X is dissociated, so in the equilibrium we have a concentration that is equal to A - X. The initial concentration of the ions were 0, and they were produced in a concentration of X. So the equilbrium concentration of H+ and F- are +X.
We said that the Ka represents the relationship between the equilibrium concentrations of the reactants and products.
Ka = [H⁺] * [F⁻]/[HF]
And we found with the ICE table that the equilibrium concentrations are:
[HF] = A - X [H⁺] = [F⁻] = x
We can replace those values in the expression of the equilibrium constant:
Ka = [H⁺] * [F⁻]/[HF] = X * X / (A-X) = X²/(A-X)
Ka = X²/( A - X)
Now we usually make another assumption. Usually X is really small when compared to A. So, we assume that X is negligible compared to A. The approximation is this one:
A - X = A
We use that approximation in our equation:
Ka = X²/( A - X ) = X²/A
And we can solve that equation for X, and find the concentration of H+ in the equilbirium.
Ka = X²/A
Ka * A = X²
[H⁺] = X = √(Ka*A)
So, to find the pH of a weak acid we use this formula:
pH = -log [H⁺]
pH = -log (√(Ka*A))
And sometimes we apply some math, and we convert that formula into this one:
pH = -1/2 * (log Ka + log A)