Ionic EquilibriumHard
Question
Calcium lactate is a salt of weak acid and represented as Ca(Lac)2. A saturated solution of Ca(Lac)2 contains 0.125 mole of salt in 0.50 L solution. The pOH of this is 5.60. Assuming complete dissociation of the salt, calculate Ka of lactate acid (log 2.5 = 0.4).
Options
A.1.25 × 10−11
B.8.0 × 10−4
C.3.2 × 10−17
D.4 × 10−5
Solution
$\left\lbrack LaC^{-} \right\rbrack_{0} = \frac{0.125}{0.5} \times 2 = 0.5\text{ M}$
Now, $P^{OH} = 7 - \frac{1}{2}\left( P^{K_{a}} + \log C \right)$
$5.6 = 7 - \frac{1}{2}\left( P^{K_{a}} + \log 0.5 \right) $$$\therefore P^{K_{a}} = 3.1 \Rightarrow K_{a} = 8 \times 10^{- 4}$$
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