Calculating the pH of the endpoint in a titration of weak acid and strong base

Show that the pH at the endpoint in the titration of $\mathrm{0.100\ mol\ dm^{-3}}$ ethanoic acid with $\mathrm{0.100\ mol\ dm^{-3}}$ sodium hydroxide is 8.72. The dissociation constant Ka of ethanoic acid is $\mathrm{1.80\ x\ 10^{-5}\ mol\ dm^{-3}}$.

I managed to obtain the value of Kb using the formula (Ka x Kb = Kw = $\mathrm{10^{-14}}$).

I worked out the chemical equation at the endpoint which is:

$$\ce{CH3COO- + H2O <--> CH3COOH + OH- }$$

However, I am unsure as to what the concentration of $\ce{CH3COO-}$ should be.

I understand that if I find [$\ce{OH-}$] then I would be able to find pOH and by extension pH.

• Would it be correct to assume that the concentration of CH3COO- to be the concentration of ethanoic acid that was present at the start of the titration (0.100 mol dm^-3), given that at the equivalence point, all CH3COOH had reacted to form a solution containing only the conjugate base CH3COO- ? – b3nj4m1n Feb 17 '17 at 10:54
• I also have a major problem with the wording of the question. The endpoint is when the end of the titration is detected like when phenolphthalein has changed pink. The equivalence point is when you have added as many moles of base as there were moles of acid in the solution. So technically the problem as stated is unanswerable. The point of course that we want to make pH(endpoint) $\approx$ pH(equivalence point ) by selecting a good indicator for the titration. – MaxW Feb 17 '17 at 17:51

Would it be correct to assume that the concentration of $\ce{CH3COO-}$ to be the concentration of ethanoic acid that was present at the start of the titration ($\mathrm{0.100~mol~dm^{-3}}$), given that at the equivalence point, all $\ce{CH3COOH}$ had reacted to form a solution containing only the conjugate base $\ce{CH3COO-}$?

You can easily rearrange the $\text{K}_\text{a}$ equation to solve for the information that you need to answer that question.

$$\text{K}_\text{a} =\frac{\ce{[H^+][A^-]}}{\ce{[HA]}}$$ or
$$\frac{\ce{[HA]}}{\ce{[A^-]}} = \frac{\ce{[H^+]}}{\text{K}_\text{a}} = 1.06\times10^{-4}$$

Since $1.06\times10^{-4}$ is insignificant compared to 0.100 you can assume that essentially all the HA is converted to $\ce{A^-}$.

NOTE: This does not mean that $\ce{[HA] = 0}$

• We can't assume [HA] = 0 because of the hydrolysis that A- undergoes in solution to form back HA. Is that correct? – b3nj4m1n Feb 17 '17 at 23:38
• Exact right. Even if you add pure A- into the solution by dissolving a salt (e.g sodium acetate), then A- undergoes hydrolysis in solution to form some HA (e.g. for sodium acetate you'd get acetic acid). – MaxW Feb 17 '17 at 23:41
• Realize that this is about significant figures. So if you add 0.100 moles of sodium acetate to a liter of distilled water you can assume that the acetate concentration is 0.100 because little acetic acid is formed. But in math class this wouldn't work because 0.1 means 0.1000000... – MaxW Feb 17 '17 at 23:57