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6-2.Equilibrium-II (Ionic Equilibrium)
medium
If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005$ $M$ calcium lactate solution at $25^{\circ}\,C$ is $........\times 10^{-1}$ (Nearest integer)

A
$85$
B
$84$
C
$83$
D
$82$
(JEE MAIN-2023)
Solution
Concentration of calcium lactate $=0.005\,M ,:$ concentration of lactate ion $=(2 \times 0.005)\,M$. Calcium lactate is a salt of weak acid $+$ strong base $\therefore$ Salt hydrolysis will take place.
$pH =7+\frac{1}{2}( pKa +\log C )$
$=7+\frac{1}{2}(5+\log (2 \times 0.005))$
$=7+\frac{1}{2}[5-2 \log 10]=7+\frac{1}{2} \times 3=8.5=85 \times 10^{-1}$
Standard 11
Chemistry