Question

The pH of a solution prepared by mixing 55.0 mL of 0.183 M KOH and 10.0 mL of 0.145 M HC2H3O2 is ________.

the answer is 13.122 however when i try to plug it into the Henderson Hasselbalch equation (pH= -log(1.8E-5)+log( 0.010065/0.00145)
i am not getting 13.122
the limiting reagent i think it is HC2H3O2
55.0 mL (1L/1000mL)(0.183mol/1L) = 0.010065 mol KOH
10.0 mL (1L/1000mL)(0.145mol/1L) = 0.00145 mol HC2H3O2

I do not know what I am doing wrong to get a pH of 5.58 instead of 13.122

can anyone tell me what I am doing wrong? I appreciate any help.
Thank You!

Answer 1

Yes, you are assuming you have buffered solution but you don't. You have an excess of KOH, which you note when you say acetic acid is the limiting regent. So (OH^-) = mols KOH/L solution. Solve for pOH and convert to pH. 13.12