Question

Adding 10.0 mL of 0.500 M HCl to 100.0 mL of pH = 7 water produces a solution with a pH of 1.34. What is the pH of the resulting solution when 10.0 mL of 0.500 M HCl is added to 100.0 mL of the buffer solution in the introductory example? That buffer contained 0.100 moles of both CH3COOH and CH3COO- in 0.100 L of water?
The Ka of acetic acid is 1.8105

Answer 1

millimols CH3COOH = 100
millimols CH3COO- = 100
millimols HCl = 0.5M x 10 mL = 5 mn.
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The added HCl will add to the base (acetatae ion) to make CH3COOH.
........CH3COO- + H+ ==> CH3COOH
I.........100......0..........100
add...............5
C.........-5.....-5..........+5
E.........95......0...........105

Substitute the E line into the HH equation and solve for pH of the final solution. Compare that with 1.34 in an un-buffered solution.