JEE Main 2023 · 24 January · evening · Numerical answer · +4 / −0
If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25°C is ___________ × 10⁻¹ (Nearest integer)
Lactic acid H; |; CH₃, —, C, —, COOH; |; OH
Answer and solution
Answer 85
Solution
Chemistry ma’am · 4 steps
Calcium lactate releases two lactate ions, so double the concentration before using the formula.
- Find the anion. Calcium lactate is Ca(CH₃CH(OH)COO)₂, a salt of a weak acid and a strong base. Only the lactate ion hydrolyses, and each formula unit releases two of them, so [lactate] = 2 × 0.005 = 0.01 M.
- Hydrolysis formula. Lactate takes a proton from water and releases OH⁻. For a salt of a weak acid and a strong base, this gives the result below, where C is the concentration of the hydrolysing anion.
pH = 7 + 1/2pKₐ + 1/2 log C
- Put in numbers. pH = 7 + 1/2(5) + 1/2 log(10⁻²) = 7 + 2.5 - 1 = 8.5.
- So. pH = 8.5 = 85 × 10⁻¹, so the answer is 85.
The common slip: 83. Using the salt concentration 0.005 M instead of the lactate concentration 0.01 M gives pH 8.349, which reads as 83, not 85.
The idea to keep. In salt hydrolysis, C is the concentration of the ion that hydrolyses, not of the salt. Check the formula: a divalent cation brings two anions with it.
Answer from the NTA final answer key. Solution by PrepWiser.