C6H5COOK(aq) yields H+(aq) + C6H5COO-(aq)
If a .045M solution of benzoic acid has an (H+)=1.7x10^-3, what is the Ka of the benzoic acid?
2007-04-08 14:59:43 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Your equation has a type, but lets call the acid Be-COOH. Then Be-COOH -> Be-COO- + H+
The equilibrium reaction is
[Be-COO-][H+]/[BeCOOH] = Ka
We start with 0.045 M BeCOOH. A certain amount of moles, x, is dissociated until equil. is attained. There,
x^2/0.045-x = Ka.
Since we know x, we can find Ka:
2.9x10-6/ 0.043= Ka
Ka= 6.7x10-5 appx.
2007-04-08 15:18:12 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
(1.7 x 10^-3)^2 divided by .045
P.S. That K is an H, right? Otherwise you have the conjugate base, and you will not produce H+.
2007-04-08 22:04:11 · answer #2 · answered by chemmie 4 · 0⤊ 0⤋