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NVAEasyJEE 2023Hydrolysis of Salts

JEE Chemistry 2023 Question with Solution

If the pKa\text{p}K_a of lactic acid is 55, then the pH\text{pH} of 0.005 M0.005 \, \text{M} calcium lactate solution at 25∘C25^\circ\text{C} is:

Answer

Correct answer:8.5

Step-by-step solution

Standard Method

Given: pKa=5\text{p}K_a = 5 and concentration of calcium lactate =0.005 M= 0.005 \, \text{M}.

Find: The pH\text{pH} of the solution.

Calcium lactate dissociates completely, so the concentration of lactate ions is

[Lactate]=2×0.005=0.01 M[\text{Lactate}] = 2 \times 0.005 = 0.01 \, \text{M}

Calcium lactate is a salt of a weak acid and a strong base, so salt hydrolysis occurs. Using the relation shown in the solution:

pH=7+12(pKa+log⁡[Lactate])\text{pH} = 7 + \frac{1}{2}(\text{p}K_a + \log[\text{Lactate}])

Substituting the values:

pH=7+12(5+log⁡(0.01))\text{pH} = 7 + \frac{1}{2}\left(5 + \log(0.01)\right)

Since log⁡(0.01)=−2\log(0.01) = -2,

pH=7+12(5−2)=7+32=8.5\text{pH} = 7 + \frac{1}{2}(5 - 2) = 7 + \frac{3}{2} = 8.5

Thus, the pH\text{pH} is 8.58.5. The solution also writes this as 85×10−185 \times 10^{-1}.

Common mistakes

  • Using the salt concentration directly as lactate ion concentration is incorrect here because each formula unit of calcium lactate gives two lactate ions. Use [Lactate]=2×0.005=0.01 M[\text{Lactate}] = 2 \times 0.005 = 0.01 \, \text{M} instead.

  • Applying a weak acid formula directly to lactic acid is wrong because the question is about its salt, calcium lactate. Treat it as a salt of a weak acid and strong base, so hydrolysis determines the pH\text{pH}.

  • A common sign error is taking log⁡(0.01)\log(0.01) as +2+2 instead of −2-2. For 0.01=10−20.01 = 10^{-2}, the correct value is −2-2.

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