A 25.00-mL sample of propionic acid, HC 3 H 5 O 2 , of unknown concentration was

Question

1. A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.181 M KOH. The equivalence point was reached when 39.72 mL of base had been added. What is the hydroxide-ion concentration at the equivalence point?  Kafor propionic acid is 1.3 × 10-5 at 25°C. A. 1.2 × 10-3M
   B. 1.0 × 10-7M
   C. 1.5 × 10-9M
   D. 9.2 × 10-6M
   E. 1.1 × 10-5M
   

1. A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.181 M KOH. The equivalence point was reached when 39.72 mL of base had been added. What is the hydroxide-ion concentration at the equivalence point?  Kafor propionic acid is 1.3 × 10-5 at 25°C. A. 1.2 × 10-3M
   B. 1.0 × 10-7M
   C. 1.5 × 10-9M
   D. 9.2 × 10-6M
   E. 1.1 × 10-5M
   

A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.181 M KOH. The equivalence point was reached when 39.72 mL of base had been added. What is the hydroxide-ion concentration at the equivalence point?  Kafor propionic acid is 1.3 × 10-5 at 25°C. A. 1.2 × 10-3M
B. 1.0 × 10-7M
C. 1.5 × 10-9M
D. 9.2 × 10-6M
E. 1.1 × 10-5M
A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.181 M KOH. The equivalence point was reached when 39.72 mL of base had been added. What is the hydroxide-ion concentration at the equivalence point?  Kafor propionic acid is 1.3 × 10-5 at 25°C. 1.2 × 10-3M
1.0 × 10-7M
1.5 × 10-9M
9.2 × 10-6M
1.1 × 10-5M
A. 1.2 × 10-3M
B. 1.0 × 10-7M
C. 1.5 × 10-9M
D. 9.2 × 10-6M
E. 1.1 × 10-5M

Explanation / Answer

The titration reaction: HC3H5O2 + KOH ==> H2O + KC3H5O2 (potassium propionate)

Moles KOH added = M KOH x L KOH = 0.181 x 0.03972 = 0.00719 moles
The balanced equation tells us that moles of KOH added = moles of KC3H5O2 formed.

The volume at the titration equivalence point is 25.00 mL + 39.72 mL = 64.72 mL = 0.06472 L.

The molarity of the propionate ion is 0.00719 / 0.06472 = 0.111 M C3H5O2-.

C3H5O2- + H2O <------> HC3H5O2 + OH-
Kb = Kw/Ka = 7.7 x 10^-10 = x^2 / [0.111-x]
x = [OH-]= 9.2 x 10^-6 M

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