Answers:
1. 282 g/mol
2. (a) pH = 10.0; (b) pKₐ = 5.5; Kₐ = 3 × 10⁻⁶; (c) thymolphthalein
Step-by-step explanation:
1. Molar mass of unknown acid
The equation for the reaction is
HA + NaOH ⟶ NaA + H₂O
(a) Calculate the moles of NaOH
Moles of NaOH = 0.023 64 L NaOH × (0.1 mol NaOH/1 L NaOH)
= 2.364 × 10⁻³ mol NaOH
(b) Moles of HA
Now, you use the molar ratio from the balanced chemical equation to find the moles of unknown acid.
Moles of HA = 2.364 × 10⁻³ mol NaOH × (1 mol HA/1 mol NaOH)
= 2.364 × 10⁻³ mol HA
(c) Molar mass of HA
You know now that 0.5632 g of HA .
MM = mass/moles = 0.5632 g/2.364 × 10⁻³ mol = 238.2 g/mol
2. Titration of acetic acid
(a) Equivalence point
The equivalence point is the pH at the steepest point of the titration curve.
In your titration of vinegar, the equivalence point appears to be at about
pH 10.0.
(b) Half-way point
At the half-way point, you have neutralized half the acid HA and converted it into the sane amount of A⁻, so [HA] = [A⁻].
Kₐ = [H₃O⁺] × [A⁻]/[HA] = [H₃O⁺]
At the half-way point, Kₐ = [H₃O⁺] and pKₐ = pH.
Your equivalence point is at about 21.3 mL, so the half-way point is at 10.7 mL.
The pH at 10.6 mL is about 5.5.
pH = pKₐ = 5.5
Kₐ = 10^(-pKₐ) = 10^(-5.5) = 3 × 10⁻⁶
(c) Appropriate indicator
An indicator is a weak acid in which the acid form HA is a different colour than the base form A⁻.
The colour change occurs when [HA] = [A⁻] and pH = pKₐ.
Your indicator should have pKₐ ≈ 10 (Kₐ≈ 10⁻¹⁰), so it changes colour at the equivalence point (pH 10).
The best indicator is thymolphthalein, because it has Kₐ = 10⁻¹⁰.
