Question

A \(35.0-\mathrm{mL}\) sample of \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with \(0.150 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of base have been added: (a) \(0 \mathrm{~mL},(\mathbf{b})\) \(17.5 \mathrm{~mL},(\mathrm{c}) 34.5 \mathrm{~mL},(\) d) \(35.0 \mathrm{~mL},\) (e) \(35.5 \mathrm{~mL},\) (f) \(50.0 \mathrm{~mL} .\)

Short answer

At various stages of the titration, we find the following pH values: (a) 0 mL of NaOH added: pH = 2.87 (initial acetic acid solution) (b) 17.5 mL of NaOH added: pH = 4.74 (half-equivalence point) (c) 34.5 mL of NaOH added: pH = 5.92 (just before the equivalence point) (d) 35.0 mL of NaOH added: pH = 8.72 (equivalence point) (e) 35.5 mL of NaOH added: pH = 12.32 (excess NaOH after equivalence point) (f) 50.0 mL of NaOH added: pH = 12.69 (excess NaOH after equivalence point)

Step-by-step solution

Write down the balanced chemical equation for the neutralization reaction between acetic acid and NaOH.

The balanced chemical equation for the reaction is: \[ \mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{3}\mathrm{COO^{-}} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Na^{+}} \]

Calculate the initial moles of acetic acid.

To get the initial moles of acetic acid, use the given volume (35.0 mL) and concentration (0.150 M): Moles of acetic acid = volume × concentration Moles of acetic acid = \(35.0 \times 10^{-3} \mathrm{L} \times 0.150 \mathrm{M}\) Moles of acetic acid = \(5.25 \times 10^{-3}\) moles

Calculate the moles of added NaOH and the moles of acetic acid and acetate ion at each stage.

For each volume of added NaOH, calculate the moles of NaOH, and then use the stoichiometry of the reaction to find the moles of acetic acid and acetate ion: (a) 0 mL of NaOH added: No reaction occurs, the solution remains as weak acetic acid. (b) 17.5 mL of NaOH added: Moles of NaOH = \(0.150 \mathrm{M} \times 17.5 \times 10^{-3} \mathrm{L}\) = \(2.625 \times 10^{-3}\) moles Since 1 mole of CH3COOH reacts with 1 mole of NaOH, the moles of CH3COOH remaining are \(5.25 \times 10^{-3} \mathrm{moles} - 2.625 \times 10^{-3} \mathrm{moles} = 2.625 \times 10^{-3} \mathrm{moles}\), and the moles of CH3COO- (acetate ion) produced are 2.625 × 10^{-3} moles. (c) - (f): Repeat the calculation for the other volumes of NaOH added.

Calculate the concentrations of acetic acid and acetate ion at each stage.

Now we want to find the concentrations of acetic acid and the acetate ion at each stage of the titration. To do this, we divide the moles of each substance by the total volume of the solution at that point: Concentration of CH3COOH = moles of CH3COOH / total volume Concentration of CH3COO- = moles of CH3COO- / total volume For example, for 17.5 mL of NaOH added (b), the total volume is 52.5 mL (35.0 mL of acetic acid + 17.5 mL of NaOH): Concentration of CH3COOH = \(2.625 \times 10^{-3}\mathrm{moles} / (52.5 \times 10^{-3}\mathrm{L}) = 0.05\mathrm{M}\) Concentration of CH3COO- = \(2.625 \times 10^{-3}\mathrm{moles} / (52.5 \times 10^{-3}\mathrm{L}) = 0.05\mathrm{M}\) Repeat this process for the other volumes of NaOH added.

Calculate the pH at each stage.

With the concentrations of acetic acid and acetate ion at each stage, we can now find the pH at each stage. For the stages before the equivalence point (when equal moles of acid and base have reacted), we can use the Henderson-Hasselbalch equation: pH = pKa + log(\(\dfrac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}\)) For CH3COOH, the pKa value is 4.74. For example, for 17.5 mL of NaOH added (b), we have: pH = 4.74 + log(\(\dfrac{0.05}{0.05}\)) = 4.74 For the stages after the equivalence point, we'll have excess NaOH. Therefore, we'll calculate the concentration of excess OH-, and then find the pOH. The pH can be found using the relationship: pH = 14 - pOH Repeat this process for the other volumes of NaOH added to find the pH at each stage.

Key concepts

Acetic Acid

Acetic acid, which has the chemical formula \( \mathrm{CH}_3\mathrm{COOH} \), is a weak acid commonly found in vinegar. It is important to recognize that acetic acid does not dissociate completely in water. This partial disassociation is what categorizes it as a weak acid.

• At any given moment, only a small fraction of acetic acid molecules are split into acetate ions \( (\mathrm{CH}_3\mathrm{COO}^-) \) and hydrogen ions \( (\mathrm{H}^+) \).
• Because of its weak nature, when you calculate the pH of an acetic acid solution, you can't assume all acetic acid turns into hydrogen ions as you would with a strong acid.

Understanding these characteristics is crucial when performing a titration with acetic acid. In acid-base titrations involving acetic acid and a strong base such as sodium hydroxide \( (\mathrm{NaOH}) \), the acid reacts with the base to form water and an acetate ion. This reaction eventually leads to titration points where calculations based on dissociation principles help determine the pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical way of relating the pH of a solution to its pKa and the ratio of the concentrations of the conjugate base to the acid. For our titration with acetic acid, this equation is particularly useful in the buffer region, which is where the amount of acetic acid is approximately equal to the amount of acetate ions.
The equation can be given as:\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]} \right) \]

• \( \text{pK}_a \) is a constant that represents the acid dissociation of acetic acid, typically 4.74 for acetic acid.
• The equation assumes that you are dealing with solutions where the acid and base pair are present, making it ideal for calculating the pH before the equivalence point in a titration.

This equation simplifies complex equilibrium calculations and is especially valuable for determining the pH when both acetic acid and its conjugate base are present in appreciable amounts. This concept allows us to calculate the pH during the titration process efficiently until we reach the equivalence point, where a different calculation method for pH is applied.

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce water and a salt. In our titration example, acetic acid \( (\mathrm{CH}_3\mathrm{COOH}) \) is neutralized by the strong base sodium hydroxide \( (\mathrm{NaOH}) \). Here's the general gist of the reaction:
The balanced chemical equation for this specific reaction is: \[ \mathrm{CH}_3\mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} + \mathrm{Na}^+ \]

• Each mole of acetic acid reacts with an equivalent mole of sodium hydroxide, converting the acid into its conjugate base, acetate ion \((\mathrm{CH}_3\mathrm{COO}^- )\), and forming water in the process.
• As you add sodium hydroxide, the concentration of acetic acid decreases while the concentration of acetate ions increases, which is key to calculating the changes in pH at each stage of the titration.

By understanding this concept, you're equipped to predict how the pH will change throughout the titration process, from start to equivalence point and beyond. These calculations are essential for accurately determining the pH at different stages of the titration.