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Question Description

I’m working on a Chemistry question and need guidance to help me study.

 questions are related to the titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. temperature is 25 oC. 

What is the initial pH of the analyte solution?

What is the pH when 6.00 mL of the KOH solution have been added? 


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Final Answer

Ka = 1.8*10^-5 = x^2 / (0.0800 -x) [rearrange and quadratic formula]

Gives x = 1.2*10^-3 M = [H+] 

So pH = -log(1.2*10^-3) = 2.92

AFTER ADDITION OF 6.00 mL KOH

mols OH- added = (0.0700 mol/L)*(0.006 L) = 4.2*10^-4 

mols H+ present in solution = (1.2*10^-3 mol/L)*(0.025 L) = 3.0*10^-5

=> mols OH- present after addition = (4.2*10^-4) - (3.0*10^-5) = 3.9*10^-4

so [OH-] = (3.9*10^-4 mols) / (0.031 L) = 0.0126 M 

therefore pOH = -log(0.0126) = 1.90 

hence pH = 14 - 1.90 = 12.1

Tate C (183)
Rice University

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