# Question b17d5

Mar 10, 2016

${K}_{a} = 4.3 \cdot {10}^{- 5}$

#### Explanation:

The first thing to notice here is that the pH of the resulting solution is lower than $7$. This should automatically tell you that the neutralization of the acid was not complete.

A complete neutralization of a weak acid with a strong base will produce a basic solution, since the conjugate base of the acid will react with water to reform some of the weak acid and produce hydroxide anions, ${\text{OH}}^{-}$, in solution.

An incomplete neutralization will result in the formation of a buffer solution that contains a weak acid, $\text{HA}$, and its conjugate base, ${\text{A}}^{-}$.

This means that the pH of the solution can be found using the Henderson - Hasselbalch equation

color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|))

Here

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{p {K}_{a} = - \log \left({K}_{a}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This is what you'll be solving for.

The balanced net ionic equation for this reaction looks like this

${\text{HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((Aq])^(-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice the $1 : 1$ mole ratios that exist between the two reactants, on one hand, and between both reactants and the conjugate base ${\text{A}}^{-}$, on the other.

This tells you that every mole of weak acid that takes part in the reaction consumes one mole of sodium hydroxide and produces one mole of conjugate base ${\text{A}}^{-}$.

Use the molarities and volumes of the two solutions to figure out how many moles of each you're mixing

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = \frac{n}{V} \implies n = c \cdot V \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${n}_{H A} = \text{0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * 35.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00700 moles HA}$

${n}_{O {H}^{-}} = {\text{0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * 25.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00500 moles OH}}^{-}$

Since you have fewer moles of hydroxide anions (delivered to the solution by the sodium hydroxide), these will be consumed completely by the reaction.

According to the aforementioned mole ratios, $0.00500$ moles of ${\text{OH}}^{-}$ will consume $0.00500$ moles of $\text{HA}$ and produce $0.00500$ moles of ${\text{A}}^{-}$.

This means that after the reaction is finished, the solution will contain

n_(OH^(-)) = "0 moles" -> color(red)("completely"color(white)(a)color(black)("consumed"))

${n}_{H A} = 0.00700 - 0.00500 = \text{0.00200 moles HA}$

${n}_{{A}^{-}} = 0 + 0.00500 = {\text{0.00500 moles A}}^{-}$

The total volume of the resulting solution will be

${V}_{\text{total" = "35.0 mL" + "25.0 mL" = "60.0 mL}}$

The concentrations of the weak acid and of the conjugate base will be

["HA"] = "0.00200 moles"/(60.0 * 10^(-3)"L") = "0.03333 M"

["A"^(-)] = "0.00500 moles"/(60.0 * 10^(-3)"L") = "0.08333 M"

Before plugging this into the H - H equation, try to predict what the $p {K}_{a}$ of the acid will be.

Notice that the pH is said o be equal to $4.77$, and that the buffer contains more conjugate base than weak acid. This should tell you that the $p {K}_{a}$ of the acid is lower than $4.77$.

Rearrange the H - H equation to solve for $p {K}_{a}$

pK_a = "pH" - log( (["A"^(-)])/(["HA"]))#

Plug in your values to get

$p {K}_{a} = 4.77 - \log \left(\left(0.08333 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(0.03333color(red)(cancel(color(black)("M}}}}\right)\right) = 4.37$

As predicted, the $p {K}_{a}$ of the acid is lower than $4.77$.

The acid dissociation constant, ${K}_{a}$, for this acid will be

${K}_{a} = {10}^{- p {K}_{a}} = {10}^{- 4.37} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4.3 \cdot {10}^{- 5} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.