Question

A student found that 53.2 mL of a 0.232M solution of NaOH was required to titrate 25.0 mL of an acetic acid solution of unknown molarity to the endpoint. What is the molarity of the acetic acid solution?

Answer 1

`"0.494 mol L"^(-1)`

Explanation:

The key to this problem is the balanced chemical equation for this neutralization reaction.

Acetic acid, `"CH"_3"COOH"`, a weak acid, will react with the hydroxide anions, `"OH"^(-)`, delivered to the reaction by the sodium hydroxide solution, to form acetate anions, `"CH"_3"COO"^(-)`, and water, according to the balanced chemical equation

`"CH"_3"COOH"_text((aq]) + "OH"_text((aq])^(-) -> "CH"_3"COO"_text((aq])^(-) + "H"_2"O"_text((l])`

As you can see, the acid and the base react in a `1:1` mole ratio. This tells that in order to get to the equivalence point, i.e. have a complete neutralization, you must mix equal numbers of moles of each reactant.

The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.

`color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))`

In this case, you will have

`n_(OH^(-)) = "0.232 mol" color(red)(cancel(color(black)("L"^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.01234 moles OH"^(-)`

This means that the acid solution must have contained `0.01234` moles of acetic acid.

The molarity of the acetic acid solution was

`["CH"_3"COOH"] = "0.01234 moles"/(25.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.494 mol L"^(-1)color(white)(a/a)|)))`

The answer is rounded to three sig figs.

It's worth noting that the pH of the solution at equivalence point will not be equal to `7`, like you have when a strong acid and a strong base neutralize each other completely.

Instead, the pH of the solution at equivalence point will be greater than `7`. That happens because the acetate anions, `"CH"_3"COO"^(-)`, will react with water to reform some of the acetic acid and produce hydroxide anions, which in turn will make the solution basic.