# How many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

## More details Given 1.00 L of a solution that is 0.100 M $C {H}_{3} C {H}_{2} C O O H$ and 0.100 M $C {H}_{3} C {H}_{2} C O {O}^{-}$ ... What is the buffer capacity of the solution? That is, how many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

Mar 19, 2017

Here's what I got.

#### Explanation:

The thing to remember about buffers is that they resist significant changes in pH because they convert strong acids or strong bases to weak acid or weak bases, respectively.

In your case, the buffer contains propionic acid, a weak acid, and the propionate anion, its conjugate base. Propionic acid has

${K}_{a} = 1.34 \cdot {10}^{- 5}$

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Consequently, the acid will have

$\text{p} {K}_{a} = - \log \left(1.34 \cdot {10}^{- 5}\right) = 4.87$

Now, when you add a strong acid, which we will represent as ${\text{H"_3"O}}^{+}$, to the buffer, the propionate anion will react with the hydronium cations and convert them to propionic acid

"CH"_ 3"CH"_ 2"COO"_ ((aq))^(-) + overbrace("H"_ 3"O"_ ((aq))^(+))^(color(blue)("strong acid")) -> overbrace("CH"_ 3"CH"_ 2"COOH"_ ((aq)))^(color(purple)("weak acid")) + "H"_ 2"O"_ ((l))" "color(darkorange)("(*)")

When you add a strong base, which we will represent as ${\text{OH}}^{-}$, to the buffer, the propionic acid will react with the hydroxide anions and convert them to propionate anions

${\text{CH"_ 3"CH"_ 2"COOH"_ ((aq)) + overbrace("OH"_ ((aq))^(-))^(color(blue)("strong base")) -> overbrace("CH"_ 3"CH"_ 2"COO"_ ((aq))^(-))^(color(purple)("weak base")) + "H"_ 2"O}}_{\left(l\right)}$

Notice that your buffer contains equal concentrations of weak acid and conjugate. As you know, the pH of a buffer that contains a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equation

color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"])))))

In your case, you will have

"pH" = "p"K_a + log( (["CH"_3"CH"_2"COO"^(-)])/(["CH"_3"CH"_2"COOH"]))

For equal concentrations of weak acid and conjugate base, you have

$\text{pH" = "p} {K}_{a} + {\overbrace{\log \left(1\right)}}^{\textcolor{b l u e}{= 0}}$

which gets you

$\text{pH} = 4.87$

Now, the problem wants you to determine how many millimoles of strong acid or strong base can be added to the buffer before a significant change in pH occurs.

A significant change in pH is usually defined as a change of $1$ unit. You know that when you add a strong acid, the pH of the buffer decreases.

Mind you, it does not decrease significantly, but it does decrease because by adding strong acid you're increasing the concentration of weak acid in solution.

So, you know that the pH must not drop below

$\text{pH"_ "strong acid" = "pH} - 1$

$\text{pH"_ "strong acid} = 4.87 - 1 = 3.87$

Let's say that $n$ represents the number of moles of strong acid that must be added in order to get the pH to $3.87$.

According to the $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ chemical equation, $1$ mole of conjugate base will consume $1$ mole of strong acid and produce $1$ mole of weak acid.

Since the buffer contains $0.100$ moles of weak acid and $0.100$ moles of conjugate base, you can say that at a pH equal to $3.87$, you will have

$0.100 - n \to$ moles of conjugate base

$0.100 + n \to$ moles of weak acid

Assuming that the volume of the solution does not change, you will have

$3.87 = 4.87 + \log \left(\frac{0.100 - n}{0.100 + n}\right)$

$\log \left(\frac{0.100 - n}{0.100 + n}\right) = 3.87 - 4.87$

This will be equivalent to

10^(log((0.100 - n)/(0.100 + n)) = 10^(-1)

which will get you

$\frac{0.100 - n}{0.100 + n} = 0.1$

$0.100 - n = 0.1 \cdot \left(0.100 + n\right)$

$0.100 - n = 0.0100 + 0.1 \cdot n$

Rearrange to solve for $n$

$n \cdot \left(1 + 0.1\right) = 0.100 - 0.0100$

$n = \frac{0.100 - 0.0100}{1 + 0.1} = 0.0818$

Since $n$ represents number of moles of strong acid that must be added to the buffer, you can say that you will need

$0.0818 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * (10^3"mmoles")/(1color(red)(cancel(color(black)("mole")))) = color(darkgreen)(ul(color(black)("81.8 mmoles}}}}$

of strong acid in order to decrease the pH of the buffer by $1$ unit.

You can use the same approach to find the number of millimoles of strong base that must be added in order to increase the pH of the buffer by $1$ unit.

Right from the start, you should be able to say that you will need $81.8$ mmoles of strong base to get the pH of the buffer from $4.87$ to $5.87$ because the weak acid and the conjugate base are initially present in equal concentrations in the buffer.