# A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?

## The pKa of acetic acid is 4.76.

May 8, 2016

${\Delta}_{\text{pH}} = 0.29$

#### Explanation:

The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.

Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.

So, the Henderson-Hasselbalch equation looks like this

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

In your case, you have acetic acid, $\text{CH"_3"COOH}$, as the weak acid and the acetate anion, ${\text{CH"_3"COO}}^{-}$, as its conjugate base. The $p {K}_{a}$ of the acid is said to be equal to $4.76$, which means that you have

"pH" = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))

The pH is equal to $5$, and so

$5.00 = 4.76 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 0.24$

This will be equivalent to

${10}^{\log} \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = {10}^{0.24}$

which will give you

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = 1.74$

This means that your buffer contains $1.74$ times more conjugate base than weak acid

$\left[\text{CH"_3"COO"^(-)] = 1.74 xx ["CH"_3"COOH}\right]$

Now, because both chemical species share the same volume, $\text{120 mL}$, this can be rewritten as

${n}_{C {H}_{3} C O {O}^{-}} / \textcolor{red}{\cancel{\textcolor{b l a c k}{120 \cdot {10}^{- 3} \text{L"))) = 1.74 xx n_(CH_3COOH)/color(red)(cancel(color(black)(120 * 10^(-3)"L}}}}$

which is

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{n}_{C {H}_{3} C O {O}^{-}} = 1.74 \times {n}_{C {H}_{3} C O O H}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

So, the buffer contains $1.74$ times more moles of acetate anions that of acetic acid.

Now, the total molarity of the buffer is said to be equal to $\text{0.1 M}$. You thus have

["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.10 M"

Once again, use the volume of the buffer to write

n_(CH_3COOH)/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) + n_(CH_3COO^(-))/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) = 0.1"moles"/color(red)(cancel(color(black)("L")))

This will be equivalent to

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{n}_{C {H}_{3} C O {O}^{-}} + {n}_{C {H}_{3} C O O H} = 0.012} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Use equations $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ and $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find how many moles of acetate ions you have in the buffer

$1.74 \cdot {n}_{C {H}_{3} C O O H} + {n}_{C {H}_{3} C O O H} = 0.012$

${n}_{C {H}_{3} C O O H} = \frac{0.012}{1.74 + 1} = \text{0.004380 moles CH"_3"COOH}$

This means that you have

${n}_{C {H}_{3} C O {O}^{-}} = 1.74 \cdot \text{0.004380 moles}$

${n}_{C {H}_{3} C O {O}^{-}} = {\text{0.007621 moles CH"_3"COO}}^{-}$

Now, hydrochloric acid, $\text{HCl}$, will react with the acetate anions to form acetic acid and chloride anions, ${\text{Cl}}^{-}$

color(red)("H")"Cl"_ ((aq)) + "CH"_ 3"COO"_ ((aq))^(-) -> "CH"_ 3"COO"color(red)("H")_ ((aq)) + "Cl"_((aq))^(-)

Notice that the reaction consumes hydrochloric acid and acetate ions in a $1 : 1$ mole ratio, and produces acetic acid in a $1 : 1$ mole ratio.

Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, this gets you

n_(HCl) = "0.300 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(6.60 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{H C l} = \text{0.001980 moles HCl}$

The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain

${n}_{H C l} = \text{0 moles} \to$ completely consumed

${n}_{C {H}_{3} C O {O}^{-}} = \text{0.007621 moles" - "0.001980 moles}$

$= {\text{0.005641 moles CH"_3"COO}}^{-}$

${n}_{C {H}_{3} C O O H} = \text{0.004380 moles" + "0.001980 moles}$

$= \text{0.006360 moles CH"_3"COOH}$

The total volume of the solution will now be

${V}_{\text{total" = "120 mL" + "6.60 mL" = "126.6 mL}}$

The concentrations of acetic acid and acetate ions will be

["CH"_3"COOH"] = "0.006360 moles"/(126.6 * 10^(-3)"L") = "0.05024 M"

["CH"_3"COO"^(-)] = "0.005641 moles"/(126.6 * 10^(-3)"L") = "0.04456 M"

Use the Henderson-Hasselbalch equation to find the new pH of the solution

"pH" = 4.76 + log( (0.04456 color(red)(cancel(color(black)("M"))))/(0.05024color(red)(cancel(color(black)("M")))))

$\text{pH} = 4.71$

Therefore, the pH of the solution decreased by

Delta_"pH" = |4.71 - 5.00| = color(green)(|bar(ul(color(white)(a/a)"0.29 units"color(white)(a/a)|)))