Question

Question #b41c8

Answer 1

Here's what I got.

Explanation:

You know that a weak monoprotic acid will only partially ionize to produce hydronium cations and the conjugate base of the acid in `1:1` mole ratios, so if you take `"HA"` to be your generic monoprotic weak acid, you can say that you have

`"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A"_ ((aq))^(-)`

Now, you know that you start with `1.75` moles of weak acid and that, at equilibrium, the solution contains `1.15` moles of weak acid.

This implies that

`"175 moles " - " 1.15 moles" = "0.60 moles"`

of weak acid have ionized to produce hydronium cations and the conjugate base of the weak acid.

This implies that, at equilibrium, the solution contains `0.60` moles of hydronium cations and `0.60` moles of `"A"^(-)` `->` this is the case because for every `1` mole of weak acid that dissociates, you get `1` mole of each product.

Use the volume of the solution to calculate the equilibrium concentrations of the species involved in the reaction.

`["HA"] = "1.15 moles"/"1.5 L" = "0.767 mol L"^(-1)`

`["H"_3"O"^(+)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)`

`["A"^(-)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)`

By definition, the acid dissociation constant, `K_a`, which uses equilibrium concentrations, is equal to

`K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])`

Plug in your values to find--I'll leave the answer without added units

`K_a = (0.40 * 0.40)/(0.767) = color(darkgreen)(ul(color(black)(0.21)))`

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.

Finally, the `"pH"` of the solution is equal to

`color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))`

In your case, the solution will have a `"pH"` of

`"pH" = - log(0.40) = color(darkgreen)(ul(color(black)(0.40)))`

The answer is rounded to two decimal places because you have two sig figs for the volume of the solution.

Now, notice that the `"pH"` of the solution is quite low. However, you can say that you're dealing with a weak acid because strong acids are characterized by the fact that they ionize completely, i.e. `1` mole of acid produces `1` mole of hydronium cations, in aqueous solution to produce hydronium cations.

In your case, the acid does not ionize completely, i.e. `1` mole of acid does not produce `1` mole of hydronium cations, so you can classify it as a weak acid despite the very low `"pH"` value of its solution.