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) $$$\therefore P^{K_{a}} = 3.1 \Rightarrow K_{a} = 8 \times 10^{- 4}$$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Which oxide is amphoteric and at the same time a powerful oxidising agent :-...Which of the following result(s) is/ are correct for the equilibrium state in a solution originally having 0.1 M-CH3COOH...At 25oC, the solubility product of Mg(OH)2 is 1.0 × 10-11. At what pH, will Mg2+ ion start precipitating in the for...Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2 ×10-4 mol L-1 Solubility product of Ag2C2O4...H3PO4 is a weak triprotic acid; approximate pH of 0.1 M Na2HPO4 (aq.) is calculated by :-...