Titration of a Weak Acid and Strong Base — Examples

We've been examining acid and base titrations. Titrations are an experimental method for finding an unknown's concentration. So far, we've studied strong acid-strong base titrations and weak base-strong acid titrations. Each one made a unique titration curve. Today, we'll examine a weak acid-strong base titration.

What's Unique about the Titration Curve

With a weak acid, the titration is a competition between two reactions. The first reaction is the acid's dissociation. The second is the neutralization of the hydrogen ion. Let's write the balanced equations for a generic weak acid.

$HA \rightleftharpoons H^{+} + A^{-}$

$HA + OH^{-} \rightleftharpoons H_{2}O + A^{-}$.

Let's explore how these two reactions affect the titration curve.

• First, we're starting with a weak acid. So, the initial pH will be higher than a strong acid. For example, hydrochloric acid could have an initial pH of 1.0. But, acetic acid may have an initial pH closer to 3.0.

• Like weak base-strong acid titrations, the curve has a tail. The tail indicates a quick initial pH change. What causes the sudden pH change? The neutralization reaction! A weak acid has a slight dissociation. So, tiny volumes of strong bases neutralize most of the hydrogen ions. Our pH increases dramatically!

• Why does the curve suddenly level off? We created a buffer! As the base neutralizes the acid, we generate the conjugate base. But, what happens to the acid's dissociation? From Le' Chateliers principle, we know it will shift toward the reactants. We're reforming some of the acid! So our pH changes slowly.

• Partway along this section, we reach the half-equivalence point. The half-equivalence point is when the $pH=pK_{a}$. So, we neutralized exactly half the acid. In the Henderson-Hasselbalch equation, $log{(\frac{[HA]}{A^{-}})}=0$. The half-equivalence is very helpful for determining the unknown. For example, what if the half equivalence occurs at a pH of 4.7? We might suspect our acid is acetic acid!

• Eventually, we overcome the acid's buffering capacity. So, we are approaching the equivalence point. Near the equivalence point, the curve becomes nearly vertical. Unlike strong acid-strong base titrations, the equivalence pH is above 7.0. Why? We have neutralized the acid. But we are left with the conjugate base! It's raising the pH above 7.0.

• Soon after the equivalence point, our curve begins to flatten. The flattening starts near a pH of 10. At this point, we've finished the titration.

Calculation Examples

In the lab, we do a titration to find the concentration of an unknown. But, when you learn about titrations, you'll have to solve problems. Let's look at 25.0 milliliters of 0.15 molar acetic acid. We'll titrate it with 0.15 molar sodium hydroxide.

Find the initial pH of acetic acid.

To solve this problem, we need the acid ionization constant. For acetic acid, it's $1.75\times10^{-5}$. For weak acids, we assume the initial concentration stays constant.

$K_{a}=\frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}\rightarrow 1.75\times10^{-5} = \frac{x^{2}}{0.15}\rightarrow x=0.0016$

$pH=-log{([H^{+}])}=-log{(0.0016)}=2.79$

Find the pH after 5.0 ml of sodium hydroxide addition.

Let's start by writing our balanced chemical equation.

$CH_{3}COOH + NaOH \rightarrow H_{2}O + CH_{3}COONa$

Our strategy is to find our initial moles of acetic acid. Then we need to know the moles of sodium hydroxide added. We know sodium hydroxide is our limiting reagent. Also, the mole ratio is 1:1:1. So we can calculate the moles of acetic acid and moles of acetate. Then we can convert that to a concentration. And finally, we can use the Henderson-Hasselbalch equation to calculate pH.

$0.15\frac{mols}{L}\times .025 L =0.00375 mols_{CH_{3}COOH}$

$0.15\frac{mols}{L}\times .005 L =0.00075 mols_{NaOH}=mols_{CH_{3}COONa}$

$[CH_{3}COOH]=\frac{.00375-.00075}{.030L}=\frac{.003mols}{.030L}=0.1\frac{mols}{L}$

$[CH_{3}COO^{-}]=\frac{.00075mols}{.030L}=0.025\frac{mols}{L}$

$pH=pK_{a} + log{(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})}\rightarrow pH= 4.76+ log{(\frac{0.025}{0.1})}=4.16$

Calculate the volume of base needed to reach the equivalence point.

The equivalence point occurs when the moles of acid and base are equal. So, we need to find our initial moles of acid. Then, we can convert that to a concentration.

$0.15\frac{mols}{L}\times .025 L =0.00375 mols_{CH_{3}COOH}=mols_{NaOH\, at\, equivalence}$

$\frac{0.00375mols}{0.15\frac{mols}{L}}=0.025 L=25.0 ml$

What is the pH at the half equivalence point?

At the half-equivlance, $pH=pK_{a}$. So, our answer is $pH=4.76$ .

What is the pH at the equivalence point?

At the equivalence point, we've neutralized all our acid. So, all our acid has become the conjugate base. We can use the $K_{b}$ value for acetate to calculate the pOH. To find $K_{b}$, we can use the self-ionization of water. Then we can convert pOH to pH.

$K_{w}=K_{a}\times K_{b}\rightarrow K_{b}=\frac{10^{-14}}{1.75\times10^{-5}}=5.71\times10^{-10}$

$[CH_{3}COO^{-}]=\frac{0.00375mols}{50.0ml}=0.075M$

$K_{b}=\frac{[OH^{-}][CH_{3}COOH]}{[CH_{3}COO^{-}]} \rightarrow 5.71\times10^{-10}= \frac{x^{2}}{0.75}\rightarrow x=8.72\times10^{-6}$

$pOH=-log{(8.72\times10^{6})}=4.05$

$pH=14-pOH=14-4.05=9.95$

(Video) 17.3b pH Calculations for Weak Acid Strong Base Titrations | General Chemistry

by Chad's Prep

In this video, Chad walks you through a weak acid/strong base titration.

He shows you how to calculate the pH...

• at the start of the titration.
• before the equivalence point.
• at the half-equivalence point.
• at the equivalence point.
• after the equivalence point.

After watching this video, you'll know how to tackle the most challenging titration problems.