# Question 135b1

Jan 30, 2016

Here's what I got.

#### Explanation:

A weak monoprotic acid labeled $\text{HX}$ will partially ionize to form the conjugate base of the acid, ${\text{X}}^{-}$, and hydronium ions, ${\text{H"_3"O}}^{+}$, according to the equilibrium reaction

${\text{HX"_text((aq]) + "H"_2"O"_text((aq]) rightleftharpoons "X"_text((aq])^(-) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

Your strategy here will be to use the $p {K}_{a}$ of the acid to find the value of the acid dissociation constant, ${K}_{a}$, then use an ICE table to help you find the equilibrium concentration of hydronium ions.

$\textcolor{b l u e}{p {K}_{a} = - \log \left({K}_{a}\right) \implies {K}_{a} = {10}^{- p {K}_{a}}}$

This will get you

${K}_{a} = {10}^{- 3.72} = 1.9 \cdot {10}^{- 4}$

The ICE table for this equilibrium will look like this

${\text{ ""HX"_text((aq]) + "H"_2"O"_text((aq]) " "rightleftharpoons" " "X"_text((aq])^(-) " "+" " "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

color(purple)("I")" " " "0.52" " " " " " " " " " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " "(+x)
color(purple)("E")" "0.52-x" " " " " " " " " " " " " " " "x" " " " " " " " " "x

You know that ${K}_{a}$ is defined as

$\textcolor{b l u e}{{K}_{a} = \left(\left[\text{X"^(-)] * ["H"_3"O"^(+)])/(["HX}\right]\right)}$

In your case, you will have

${K}_{a} = \frac{x \cdot x}{0.52 - x} = {x}^{2} / \left(0.52 - x\right) = 1.9 \cdot {10}^{- 4}$

Because ${K}_{a}$ has such a small value compared with the initial concentration of the weak acid, you can use the following approximation

$0.52 - x \approx 0.52$

This will get you

$1.9 \cdot {10}^{- 4} = {x}^{2} / 0.52$

Solve for $x$ to find

$x = \sqrt{0.52 \cdot 1.9 \cdot {10}^{- 4}} = 9.94 \cdot {10}^{- 3}$

Since $x$ represents the equilibrium concentration of the conjugate base and of the hydronium ions, you will have

$\left[{\text{H"_3"O}}^{+}\right] = x = 9.94 \cdot {10}^{- 3}$

The pH of the solution will thus be

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

$\text{pH} = - \log \left(9.94 \cdot {10}^{- 3}\right) = \textcolor{g r e e n}{2.0}$

Now, what happens when half of the acid is neutralized by a strong base like sodium hydroxide?

The acid will react with the base, which I'll represent as ${\text{OH}}^{-}$, to form water and ${\text{X}}^{-}$, the conjugate base of the acid

${\text{HX"_text((aq]) + "OH"_text((aq])^(-) -> "X"_text((aq])^(-) + "H"_2"O}}_{\textrm{\left(a q\right]}}$

Now, for every mole of strong base added to the acid solution, one mole of weak acid will be consumed and one mole of conjugate base will be produced.

Neutralizing half of the acid is equivalent to converting half of its moles to moles of water and moles of conjugate by adding strong base.

So if $n$ represents the total number of moles of weak acid, you will need $\frac{n}{2}$ moles of sodium hydroxide to consume $\frac{n}{2}$ moles of $\text{HX}$.

This will of course produce $\frac{n}{2}$ moles of ${\text{X}}^{-}$.

The resulting solution will contain $\frac{n}{2}$ moles of $\text{HX}$ and $\frac{n}{2}$ moles of ${\text{X}}^{-}$. Since the volume is the same for both species, their concentrations will be identical.

$\textcolor{b l u e}{\left[{\text{HX"] = ["X}}^{-}\right]} \to$ when half of the acid has been neutralized

Plug this into the expression for ${K}_{a}$ to get

K_a = ( color(red)(cancel(color(black)(["X"^(-)]))) * ["H"_3"O"^(+)])/color(red)(cancel(color(black)(["HX"])))

${K}_{a} = \left[{\text{H"_3"O}}^{+}\right]$

Therefore,

"pH" = - log(["H"_3"O"^(+)]) = overbrace(- log(K_a))^(color(purple)(pK_a))#

$\text{pH} = \textcolor{g r e e n}{p {K}_{a}}$

And there you have it. When equal concentrations of weak acid and conjugate base are present in solution, the pH of the buffer is equal to the acid's $p {K}_{a}$.