Ionic Equilibrium Question 2
Question 2 - 24 January - Shift 2
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)
Lactic acid $CH_3-\stackrel{H}{O}-COOH$
Show Answer
Answer: (85)
Solution:
Formula: Salt hydrolysis of weak acid and strong base
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.
$ \begin{aligned} & 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} \end{aligned} $[^2]
In resultant solution
$ n _{NH_3}=0.1-0.02=0.08 $
$ n _{NH_4 Cl}=n _{NH_4^{+}}=0.1+0.02=0.12 $
$pOH=pK_b+\log \frac{[NH_4^{+}]}{[NH_3]}$
$ \begin{aligned} & =4.745+\log \frac{0.12}{0.08} \\ & =4.745+\log \frac{3}{2} \\ & =4.745+0.477-0.301 \\ & pOH=4.921 \\ & pH=14-pH \\ & \quad=9.079 \end{aligned} $
