Acid dissociation constant

Problem:



Solution:

Let's determine what are the chemicals first.

$CH_3CO_2H$ is acetic acid, it is a weak acid in aqueous solution.

$NaCH_3CO_2$ is sodium acetate, it is a salt.

$HCl$ is hydrochloric acid, of course.

First of all, we have 100 mL = 0.1 L of the buffer solution, so we have

$0.2 \times 0.1 = 0.02$ mol of $CH_3CO_2H$ and
$0.34 \times 0.1 = 0.034$ mol of $NaCH_3CO_2$.

The salt is an ionic compound, so it dissolves in water and become $Na^+$ and $CH_3CO^-_2$

The fact that acetic acid is a weak acid means we need to compute how well it dissociate in water using the constant, in particular, we have these two equations

$pK_a = -\log K_a$
$K_a = \frac{[A^-][H^+]}{[HA]}$

With the first formula, we can solve for $K_a = 10^{-4.74}$.

Now we have the reversible reaction:

$CH_3COOH \to  CH_3COO^- + H^+$, let say, out of 0.02 mol of $CH_3COOH$, $p$ mol of them got dissociated, so we get the total number of moles of elements as:

$\begin{eqnarray*} [CH_3COOH] &=& 0.02 - p \\ [CH_3COO^-] &=& p + 0.034 \\ [H^+] &=& p \\ [Na^+] &=& 0.034 \end{eqnarray*}$

Plug these values into the $K_a$ formula, we can solve for $p$ as follow:

$10^{-4.74} = \frac{(p + 0.034)(p)}{0.02-p}$

The resulting quadratic formula has two roots, rejecting the obviously wrong negative solution, we have solved $p = 1.0695 \times 10^{-5}$.

So the pH value is $-\log p = 4.97$.

Now we add hydrochloric acid. We added 8ml = 0.008L of 1M $HCl$, so we have added $0.008 \times 1 = 0.008$ mol of $HCl$.

Hydrochloric acid is a strong acid and we assume it completely dissociate (not completely true, but a good approximation), so the number of moles changes as follow:

$\begin{eqnarray*} [CH_3COOH] &=& 0.02 - p \\ [CH_3COO^-] &=& p + 0.034 \\ [H^+] &=& p + 0.008 \\ [Na^+] &=& 0.034 \\ [Cl^-] &=& 0.008 \\ \end{eqnarray*}$

Now we solve the $pK_a$ formula again and get $10^{-4.74} = \frac{(p + 0.034)(p + 0.008)}{0.02-p}$, but wait, the formula only generate negative $p$. Now, what does it mean? It means instead of dissociating, the acetate ions take the proton and form the acid instead! In this case, we have $p = -0.00798$ and therefore the proton concentration is $-0.00798 + 0.008 = 1.95684 \times 10^{-5}$, so the pH is $-\log (p + 0.008) = 4.71$.

As expected, it becomes more acidic!