Can assumptions made during calculation of ka of buffer solutions be justified?

Question

We make many assumptions in sciences which are not always true but as they make little difference we ignore them for simplicity. But today I have decided that I can no longer be lured into using expression of $K_{a}$ and $K_{w}$ by assumptions which I deem to make a significant difference.

1. We say that $[H_{2}O]$ is constant. Why I think it is not justified?
   $$\ce{H2O + HCL <=> H3O+ + Cl-}$$ $[H_{2}O]$ is almost $ 55.6 \:\mathrm{mol\:dm^-3}{}$.
   At around $20^{o}C$ $20\:\mathrm{mol}$ of $HCl_{aq}$ can be dissolved. Let's say we dissolve $20 \:\mathrm{mol}$ of $HCl$ and try to find it's WE ASSUME that % disassociation of water is $100 \%$. So we must take $[HCl]\approx0$ that means that $20\:\mathrm{mol}$ of $H_{2}O$ has reacted and $[H_{2}O]$ is no longer a constant but has rather changed to $35.6\mathrm{mol}$.
   So this expression can no longer be used. $$K_{a}=\frac{[H_{3}O^{+}].[Cl^{-}]} {[HCl]}=K_{c}.[H_{2}O]$$ Note: you might point out that $K_{a}$ is not used for strong acids but I just used $HCl$ as an example but make my point easier to explain.
2. Assumption made about concentrations when calculating $Ka$ of buffer solutions.
   $$CH_{3}COOH_{aq} \rightleftharpoons CH_{3}COO^{-}_{aq} + H^{+}_{aq}$$ We have added $CH_{3}COONa$ to make a buffer solution.
   $$CH_{3}COONa_{aq} \rightarrow CH_{3}COO^{-}_{aq} + Na^{+}_{aq}$$ So,
   $$[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa} + [CH_{3}COO^{-}]_{CH_{3}COOH}$$ $$[CH_{3}COO^{-}]_{CH_{3}COOH} = [H^{+}_{aq}]$$ BUT WE ASSUME that $[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa}$ because we say $[CH_{3}COO^{-}]_{CH_{3}COOH}$ is negligible yet we include [$H_{aq}^{+}]$ in the expression which has same value as $[CH_{3}COO^{-}]_{CH_{3}COOH}$. Same value is considered negligible at one place and not negligible at another in the same expression. How is that justified?

Answer 1

Answering your first question, $[\ce {H2O}]$ is constant because the square brackets indicate a concentration. Concentration is the number of moles of the solute on the volume of the solvent.

Except in this case, water is the solvent. So the concentration will always be constant, since the volume is proportional to the number of moles.

Besides, while dealing with the law of mass action, for liquids we don't take concentration, we take activity. Which is approximately one for water.

As for your second question, we basically have the following product:

$$\ce{([CH3COO- ]_{CH3COOH} + [CH3COO- ]_{CH3COONa})[H+ ]}$$

which can be rewritten as $(a + b)b$, where $b << a$. Let's say $a$ was $1$ and $b$ was $0.0001$.

The actual value of the expression is $1.0001\times 10^{-4}$. Neglecting only the first $b$ will give $1\times 10^{-4}$, which is not too much different.

Neglecting only the second $b$ will give $1.0001$, and neglecting both will give $1$. Both of these answers are very wrong.

When a small number is added to a large one, the small number can be ignored. But not when it is multiplied.

Answer 2

For the first part I was forgetting this simple this: $$[H_{2}O]=\frac{n(H_{2}O)}{Volume_{H_{2}O}} mol.dm^{-3}$$ So as the $n(H_{2}O)$ will decrease the $Volume_{H_{2}O}$ will decrease too making $[H_{2}O]$ a constant.
The second part have been answered by ManishEarth.