Why wouldn't you consider the [H+] produce from the weaker acid in a mixture of two acid?

Question

Calculate the pH of a solution that contains $\pu{2.0 M}$ $\ce{HF}$ and $\pu{2.0 M}$ $\ce{C6H5OH}$. Also calculate the concentration of $\ce{C6H5O-}$ in this solution at equilibrium.

In these situation, you are told to first find which is the strongest acid. $\ce{HF}$ is.

Knowing this, you can use the ICE table and the $K_a$ of $\ce{HF}$ ($7.2 \times 10^{-4}$) to find $[\ce{H+}]$ in the dissociation of $\ce{HF}$. This $[\ce{H+}]$ is then use to find the $\mathrm{pH}$ of the mixture, but my question is why wouldn't we add the $[\ce{H+}]$ produced from the dissociation of $\ce{C6H5OH}$? Also the $[\ce{H+}]$ that could be made from $\ce{H2O}$ reacting with itself? Is it because the $[\ce{H+}]$ from $\ce{C6H5OH}$ (weaker acid) and $\ce{H2O}$ is so low that it won't really affect the $\ce{pH}$ number?

This question pours over to when you calculate $[\ce{C6H5O-}]$. Why would we just substitute the $[\ce{H+}]$ we found from $\ce{HF}$ for the $[\ce{H+}]$ in the $K_a$ formula of the dissociation of $\ce{C6H5OH}$:

$$K_a = \frac{[\ce{H+}][\ce{C6H5O-}]}{2.0 - x}?$$

Is it because at equilibrium of the mixture we know $[\ce{H+}]$ must be close to the $[\ce{H+}]$ that would be produce at the equilibrium of a $\ce{HF}$ reaction? So it will somehow affect which way the $\ce{C6H5OH}$ reaction will go (Le Chatelier's principle). So because of the addition of HF $[\ce{H+}]$, the reaction will go left and less $\ce{C6H5O-}$ should be made compare if it was just a solution of $\ce{C6H5OH}$?

Answer 1

You've practically answered it yourself... The $\ce{H^+}$ released from $\ce{HF}$ will be so much that it will cause an effect called the common ion effect, which is basically Le Chatelier's rule applied to ions in equilibrium. This effect will cause a reduction in the total amount of $\ce{H^+}$ that dissociates out of the weaker acid, $\ce{C6H5OH}$, due to which the amount of $\ce{H^+}$ that dissociates out of $\ce{C6H5OH}$ will be negligibly small compared to that produced from $\ce{HF}$. The same thing happens to $\ce{H2O}$. Thus, in the pH calculation, we only substitute the concentration of $\ce{HF}$, so that we can get the answer with least trouble. In reality, even the smallest quantity that dissociates out should be considered, but to make our calculations simple, we use the most significant amount, leaving out the others.

In the equilibrium equation, the same rule applies, giving us an $\ce{[H^+]}$ value same as that of $\ce{[HF]}$. Also, the $x$ in the denominator is negligibly small, so even that is neglected in the calculations.

Answer 2

You don't have to consider the system's sub-components separately, but it makes the algebra much easier.

If pKa of $\ce{HF}$ is 3.14 and pKa of $\ce{PhOH}$ is 9.95 and the problem gives you concentrations to ONE SIGNIFICANT figure, it isn't clear to me why you think the difference between .00072 and .00072000011 is worth worrying about...It can be ignored, clearly. It's not a question of the weaker acid, its a question of whether the values are close enough so that the smaller is significant with regard to the larger. In this case, clearly not. $\ce{PhOH -> PhO- + H+}$ is the chemical equation for the 2nd part, so

$$ \ce{\frac{[PhO][H]}{[PhOH]}} = K_a $$

is the equation you derive from it (ignoring difference between concentration and activity, and ignoring water).

The fact that $K_a$ for this is $10^{-10}$ and that $\ce{[H+]}$ is on the order of 0.0007 means you get $\frac{X(0.0007)}{(2-X)}=10^{-10}$ which if you have any done a minimal amount of algebra practice/drill immediately suggests that X is going to be on the order of 3E-7 and again is there really a difference between 2 and 2-0.0000003? You'll note I'm not asking the slightly different question about whether the difference between 2.0000000 and 2.0000000-0.0000003 = 1.9999997; 2 is not the same as 2.0000000 when significant figures are in play.