Chemistry

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!

asked by Lubna
  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

    posted by DrBob222

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