# Question b41c8

Dec 6, 2017

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 $\text{HA}$ to be your generic monoprotic weak acid, you can say that you have

${\text{HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A}}_{\left(a q\right)}^{-}$

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

$\text{175 moles " - " 1.15 moles" = "0.60 moles}$

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 ${\text{A}}^{-}$ $\to$ 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

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

${K}_{a} = \frac{0.40 \cdot 0.40}{0.767} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.21}}}$

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

Finally, the $\text{pH}$ of the solution is equal to

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

In your case, the solution will have a $\text{pH}$ of

$\text{pH} = - \log \left(0.40\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.40}}}$

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

Now, notice that the $\text{pH}$ 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 $\text{pH}$ value of its solution.