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The next 11 questions are related to the titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. Assume that the temperature is 25 oC.
What is the pH when 6.00 mL of the KOH solution have been added? What is the pH when 35.00 mL of the KOH solution have been added?

Feb 19th, 2015

25 ml * 1 L/1000 mL * 0.08 mol/L = 0.002 mol Acid

6 mL * 1 L/1000 mL *0.07 mol/L = 0.00042 mol KOH

35 mL * 1 L/1000 mL *0.07 mol/L = 0.00245 mol KOH

Since acetic acid is monoprotic, it will react with KOH in a 1:1.

1.  Acid is in excess

0.002-0.00042 mol = 0.00158 mol Acid / (6+25 ml) * 1000 mL/L = 0.051 M acid

Using the Ka value for Acetic acid (not given here), you could calculate the concentration of H+ ions.  remember that pH = - log[H+]

2.  Base in excess

0.00245-0.002 mol = 0.00045 mol KOH = 0.00045 mol OH- / (25+35 ml) * 1000 mL/L = 0.0075 M OH-

pOH = -log(0.0075) = 2.12

pH = 14 - pOH = 14 - 2.12 = 11.88

Feb 19th, 2015

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Feb 19th, 2015
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Feb 19th, 2015
Dec 4th, 2016
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