What is the Ka of this acid?
A saturated solution of Caproic acid ( C5H11COOH) contains 11g/L of solution and has a pH of 2.94. What is the Ka ?
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To solve this problem we have to follow these steps:
1. First we have to let the volume of a solution be 1L. So in1 L of solution there are 11 grams of Caproic acid.
2. Get the number of moles of Caproic acid.
11grams Caproic acid x (1moles caproic acid/116.16 gmol-1)
= 0.09469696969 moles Caproic acid
3. Get the moles of H+ from the pH of the solution.
pH = -log(H+)
[H+] = 10^(-pH)
[H+] = 10^(-2.94)
[H+] = 0.00114815362
4. The expression for can be derived from the dissociation of Caproic acid. We let Caproic acid = HA, a monoprotic weak acid.
HA <---> H+ + A-
ka = [H+][A-]/[HA]
[H+] = [A-] = 0.0011482
[HA] = 0.094697
ka = (0.0011482)(0.0011482)/0.094697
ka = 0.00001392078 = 1.39x10^5
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