Question

Calculate the \(\mathrm{pH}\) at the equivalence point in titrating \(0.100 \mathrm{M}\) solutions of each of the following with \(0.080 \mathrm{M} \mathrm{NaOH}\) : (a) hydrobromic acid (HBr), (b) chlorous acid (HClO \(_{2}\) ), (c) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\).

Short answer

At the equivalence point, the pH for the titration of 0.100 M solutions of each acid with 0.080 M NaOH is: (a) Hydrobromic acid (HBr): pH ≈ 12.36, (b) Chlorous acid (HClO₂): pH ≈ 1.20, and (c) Benzoic acid (C₆H₅COOH): pH ≈ 2.03.

Step-by-step solution

Identify each acid as strong or weak

We will begin by determining whether each given acid is strong or weak. (a) Hydrobromic acid (HBr) is a strong acid. (b) Chlorous acid (HClO2) is a weak acid. (c) Benzoic acid (C6H5COOH) is a weak acid. Now that we have identified the acid strength, we can proceed with finding the pH at the equivalence point.

Calculate the moles of acid and base

To calculate the moles of the acid and the base, we will use the formula: moles = concentration × volume For all three acids, let's assume that we started with a 50 mL (0.050 L) solution of 0.100 M acid. The moles of acid can be calculated as follows: moles (acid) = 0.100 M × 0.050 L = 0.005 mol Now, let's find the volume of NaOH needed to reach the equivalence point for each acid (at equivalence point, moles of acid = moles of base): moles (NaOH) = 0.005 mol Volume (NaOH) = moles (NaOH) / concentration (NaOH) = 0.005 mol / 0.080 M ≈ 0.0625 L

Calculate the pH at equivalence point for each acid

(a) For HBr, since it's a strong acid, it will completely dissociate: HBr + OH⁻ → Br⁻ + H₂O At equivalence point: [HBr] = 0 [OH⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M pOH = -log(0.044) pH = 14 - pOH ≈ 12.36 (b) For HClO2, the reaction with NaOH will produce its conjugate base ClO2⁻: HClO2 + OH⁻ → ClO2⁻ + H2O At equivalence point, all of the HClO2 will be converted to ClO2⁻: [ClO2⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M Now we have to use the Ka expression for HClO2 to calculate the [H+] concentration: Ka = [H+][ClO2⁻]/[HClO2] ≈ 1.1 × 10⁻² [H+] = sqrt(Ka × [ClO2⁻]) ≈ 6.371 × 10⁻² M pH = -log(6.371 × 10⁻²) ≈ 1.20 (c) For C6H5COOH, the reaction with NaOH will produce its conjugate base C6H5COO⁻: C6H5COOH + OH⁻ → C6H5COO⁻ + H₂O At equivalence point, all of the C6H5COOH will be converted to C6H5COO⁻: [C6H5COO⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M Now we have to use the Ka expression for C6H5COOH to calculate the [H+] concentration: Ka = [H+][C6H5COO⁻]/[C6H5COOH] ≈ 6.5 × 10⁻⁵ [H+] = sqrt(Ka × [C6H5COO⁻]) ≈ 1.034 × 10⁻² M pH = -log(1.034 × 10⁻²) ≈ 2.03 So, the pH at the equivalence point for each of the acids is: (a) Hydrobromic acid (HBr): pH ≈ 12.36 (b) Chlorous acid (HClO2): pH ≈ 1.20 (c) Benzoic acid (C6H5COOH): pH ≈ 2.03

Key concepts

Strong acids

Strong acids are fascinating because they completely dissociate in water. This means that when you dissolve a strong acid in water, every molecule will break apart into its ions. For example, hydrobromic acid (HBr) is a strong acid. When it is dissolved in water, it dissociates into hydrogen ions (H⁺) and bromide ions (Br⁻).

Characteristics of Strong Acids:

• They completely ionize in water.
• They are typically identified by their low initial pH (before titration).
• The equivalence point in their titration occurs at a high pH due to the excess of hydroxide ions.

In the case of strong acids, like HBr, during a titration with a strong base like NaOH, the pH at the equivalence point will be greater than 7. This is because the strong base will dominate, leaving the solution with more hydroxide ions.

Weak acids

Weak acids only partially dissociate in water, which means not all of their molecules break into ions. Two examples of weak acids from our exercise are chlorous acid (HClO₂) and benzoic acid (C₆H₅COOH). They behave differently from strong acids in a titration.

Characteristics of Weak Acids:

• Partial ionization leads to an equilibrium between the undissociated acid and ions.
• Their pH values are typically higher than those of strong acids before titration starts.
• The equivalence point in their titration occurs at a pH lower than 7 due to the formation of a conjugate base.

During a titration with a strong base, the weak acid is converted into its conjugate base once it reaches its equivalence point. This conjugate base then influences the pH of the solution. The solution's pH depends on the strength of the conjugate base, which may be calculated using its ionization constant, Ka.

Titration calculations

Titration is a process used to determine the concentration of a solution, either an acid or base. It involves the gradual addition of a titrant of known concentration to a solution with a substance of unknown concentration, called the analyte. When the amount of titrant added equals the amount needed to completely react with the analyte, this point is called the equivalence point.

Key Steps in Titration Calculations:

• Identify whether the substances involved are strong or weak acids/bases.
• Calculate moles of acid and base using concentration and volume: \( \text{moles} = \text{concentration} \times \text{volume}\).
• Use stoichiometry to determine the exact point of neutralization, which is the equivalence point.
• At the equivalence point, calculate the resulting pH by considering the remaining ions in the solution beyond neutralization.

For strong acids titrated with strong bases, the excess of hydroxide ions means the solution will have a pH greater than 7. For weak acids, like in our problem, the pH is influenced by the conjugate base, leading to a different calculation process involving equilibrium constants.