Question

If `53` `mL` of a base solution having an `OH^-` ion concentration of `2.35 * 10^-3` `M` is titrated with `57` `mL` of an acid (assume a strong monoprotic acid) what is the concentration of the acid (concentration of the hydronium ion)?

Answer 1

`2.8 * 10^(-3)"M"`

Explanation:

A neutralization reaction will take place when the hydronium cations produced by the acid in solution will react with the hydroxide anions produced by the base to form water.

`"OH"_text((aq])^(-) + "H"_3"O"_text((aq]) -> 2"H"_2"O"_text((l])`

Notice that you need one mole of hydronium cations to neutralize one mole of hydroxide anions. This basically means that the reaction will always consume equal numbers of moles of hydronium cations and hydroxide anions.

Judging from the volume of the two solutions used for this reaction, you can predict that the molarity of the hydronium cations will be very similar to that of the hydroxide anions.

So, use the volume an molarity of the hydroxide anions to determine how many moles you have in solution

`color(blue)(c = n/V implies n = c * V)`

`n_(OH^(-)) = 2.35 * 10^(-3)"M" * 53 * 10^(-3)"L" = 1.2455 * 10^(-4) "moles OH"^(-)`

This of course means that the sample of acid used must have contained `0.12455` moles of hydronium cations

`n_(H_2O^(+)) = 1.2455 * 10^(-4)"moles"`

The molarity of the hydronium ions will thus be

`["H"_3"O"^(+)] = (1.2455 * 10^(-4)"moles")/(57 * 10^(-3)"L") = 2.18 * 10^(-3)"M"`

If you take `"HA"` to be your generic strong acid, you can say that

`["H"_3"O"^(+)] = ["HA"] = 2.18 * 10^(-3)"M"`

That is the case because a strong acid dissociates in a `1:1` mole ratio with the hydronium ions

`"HA"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "A"_text((aq])^(-)`

Rounded to two sig figs, the number of sig figs you have for the volumes of the two solutions, the answer will be

`["H"_3"O"^(+)] = ["HA"] = color(green)(2.2 * 10^(-3)"M")`