In: Chemistry
Determine the pH of the following solutions.
(a) a 0.48 M CH3COOH solution
(b) a solution that is 0.48 M CH3COOH and 0.22
M CH3COONa
a. CH3COOH -----------------> CH3COO^- (aq) + H^+ (aq)
I 0.48 0 0
C -x +x +x
E 0.48-x +x +x
ka = [CH3COO^-][H^+]/[CH3COOH]
1.8*10^-5 = x*x/0.48-x
1.8*10^-5*(0.48-x) = x^2
x = 0.00293
[H^+] = x = 0.00293M
PH = -log[H^+]
= -log0.00293
= 2.5331
b. [CH3COOH] = 0.48M
[CH3COONa] = 0.22M
PH = Pka + log[CH3COONa]/[CH3COOH}
= 4.75 + log0.22/0.48
= 4.75-0.3388
= 4.4112 >>>answer