Calculate the pH of a 0.025 M solution of acetaldehyde (CH3CHO) at 25 °C, given that its α‑hydrogen acidity constant Ka = 1.0 × 10⁻¹⁷.
Answer
pH ≈ 9.30
The correct answer is pH ≈ 9.30 — Calculate pH: pH = –log10(5.0×10⁻¹⁰) = 9.30 (to two decimal places).
Step-by-step solution
- Write the equilibrium for deprotonation of the α‑hydrogen: CH3CHO + H2O ⇌ CH2CHO⁻ + H3O⁺.
- Let the degree of dissociation be x = [H3O⁺] = [CH2CHO⁻].
- Express Ka: Ka = (x·x)/(c – x) ≈ x² / c because x ≪ c.
- Substitute the values: x = √(Ka·c) = √(1.0×10⁻¹⁷ × 0.025) = √(2.5×10⁻¹⁹) = 5.0×10⁻¹⁰ M.
- Calculate pH: pH = –log10(5.0×10⁻¹⁰) = 9.30 (to two decimal places).
Common mistake: Using the carbonyl Ka instead of the α‑hydrogen Ka, which leads to a much lower pH.
Concept tested: Acidity of α‑hydrogen in aldehydes
Published · How this solution was prepared and checked
Related questions
- 1 2‑Propanol (CH3CHOHCH3) is oxidised by sodium dichromate (Na2Cr2O7·2H2O) in acidic medium to give acetone. If 5.0 g of 2‑propanol are used, how many moles of acetone will be formed assuming quantitative conversion?
- 2 Explain why acetone (CH3COCH3) gives a positive iodoform test when treated with aqueous iodine and sodium hydroxide. Write the step‑wise mechanism, including all intermediates, and state the final product formed.
- 3 Which of the following products is obtained when cyclohexanone is treated with sodium borohydride (NaBH4) in ethanol?
- 4 Methylamine (CH_3NH_2) is a weak base with K_b = 4.4 × 10^-4. Calculate the pH of a 0.10 M aqueous solution of methylamine at 25°C.
- 5 Calculate the boiling point elevation when 5.00 g of glucose (C_6H_12O_6, M = 180.16 g mol^-1) is dissolved in 250 g of water. The ebullioscopic constant K_b of water is 0.512 K kg mol^-1.
- 6 A solution is prepared by dissolving 0.0100 g of an unknown polymer in 20.0 mL of water at 298 K. The measured osmotic pressure of the solution is 0.085 atm. Assuming the polymer behaves as an ideal dilute solution, calculate its molar mass.