CH3CO2H → CH3CO2− + H+
[CH3CO2−] [H+] / [CH3CO2H] = 1.8E-5
x^2 / (.25 - x) = 1.8E-5 ... x^2 + 1.8E-5 x - 4.5E-6 = 0 ... x = 2.11E-3
log(2.11E-3) = 2.67 ... good choice
What is the pH of a 0.25 M acetic acid solution if the Ka of acetic acid is 1.8 x 10-5?
2.67
0.60
5.34
11.33
I got A, 2.67. I think this is correct as well since acetic acid is a weaker acid and the ph # would make sense.
2 answers
Scott you a Dawg brother