4. Calculate the percent ionization of 7.5×10 3 M butanoic acid ( Ka =1.5×10 5 )

Question

4. Calculate the percent ionization of 7.5×103M butanoic acid (Ka=1.5×105)

   [A] 2.5 [B] 3.5 [C] 4.5 [D] 5.5 [E] 6.5

5. The Ka of benzoic acid is 6.30 x10-5. The pH of a buffer prepared by combining 50.0 mL of

1.00 M potassium benzoate and 50.0 mL of 1.00 M benzoic acid is ---

[A] 3.406 [B] 1.705 [C] 0.851 [D] 2.383 [E] 4.201

6. Calculate the percent ionization of formic acid (HCO2H) in a solution that is 0.253 M in

formic acid and 0.111 M in sodium formate (NaHCO2). The Ka of formic acid is 1.77 x 10-4.

[A] 3.387 [B] 30.5 [C] 13.4 [D] 0.1621 [E] 5.05 x 10-4

Explanation / Answer

4.

%ion = [H+]/M*100

Ka = [H+][A-]/[HA]

1.5*10^-5 = x*x/((7.5*10^-3)-x)

x = 3.27*10^-4

[H+]= 3.27*10^-4

%ion = [H+]/M*100 = (3.27*10^-4)/(7.5*10^-3)*100 = 4.6%

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