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QJEEM23Ch359f: JEE Main 2023 Chemistry - Class 11 JEEM23Ch359f Solution: If the pKa of lactic acid is 5, then the pH of 0.005 M calcium…

If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25 ∘^\circ C is ___________ ×\times 10 −1^{-1} (Nearest integer)

JEE Main 2023 (Online) 24th January Evening Shift Chemistry - Ionic Equilibrium Question 26 English

Step-by-Step Verified Solution

Concentration of calcium lactate =0.005M=0.005 \mathrm{M} , concentration of lactate ion =(2×0.005)M=(2 \times 0.005) \mathrm{M} .

Calcium lactate is a salt of weak acid ++ strong base

∴\therefore Salt hydrolysis will take place.

amp;pH=7+12(pKa+log⁡C)amp;=7+12(5+log⁡(2×0.005))amp;=7+12[5−2log⁡10]=7+12×3=8.5=85×10−1\begin{aligned}& \mathrm{pH}=7+\frac{1}{2}(\mathrm{pKa}+\log \mathrm{C}) \\\\& =7+\frac{1}{2}(5+\log (2 \times 0.005)) \\\\& =7+\frac{1}{2}[5-2 \log 10]=7+\frac{1}{2} \times 3=8.5=85 \times 10^{-1}\end{aligned}

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