Question
Mols benzoic acid = 12.2g/122.12g = 0.0999
Int. Concentration of Benzoic acid = 0.0999 mol/ 0.500L = 0.1998 M
Ka = 6.3 x 10-5 = (x)(x)/0.1998 – x
X = [H+] = * M
pH = -log(*) = *
Sorry, this question is just really confusing me..
Int. Concentration of Benzoic acid = 0.0999 mol/ 0.500L = 0.1998 M
Ka = 6.3 x 10-5 = (x)(x)/0.1998 – x
X = [H+] = * M
pH = -log(*) = *
Sorry, this question is just really confusing me..
Answers
DrBob222
It takes time to type all this stuff. I was working on my response (below at the original post) all this time.