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You have just made up a solution that you made by combining 100 ml 0.1 M sodium acetate...

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mapeps | Student, Undergraduate | (Level 1) Honors

Posted September 5, 2011 at 12:04 PM via web

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You have just made up a solution that you made by combining 100 ml 0.1 M sodium acetate and 1 ml 10 M acetic acid.

 

If the pKa for dissociation of CH3COOH = 4.78, what is the pH of your solution?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 5, 2011 at 1:59 PM (Answer #2)

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When sodium acetate and acetic acid are added to make up the solution, the sodium acetate gives Na+ and Ac- ions and the acetic acid gives H+ and Ac- ions.

100 mL of 0.1 M sodium acetate gives 100*.1/1000 = 0.01 moles of Na+ and Ac- ions. 1 mL of 10 M acetic acid gives 0.01 mole of H+ and 0.01 mole of Ac- ions. The total concentration of Ac- is 0.02

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = -log(10)[Ka] = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

4.78 = log(10)[0.01] - log(10)[H+] - log(10)[0.02] => -log(10)[H+] = 4.78 + log(10)[0.02] - log(10)[0.01] => pH = 5.0810 The required pH of the solution is 5.0810

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