Question

1. Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations.

Part A 2.06×10−2 M .

2. Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 8.0×10−2 MNaOH.

Part A hydrobromic acid (HBr)

Part B chlorous acid (HClO2)

Part C benzoic acid (C6H5COOH)

Answer 1

1. ka of propionic acid = 1.34*10^-5

   Ka =cx^2

(1.34*10^-5) = (2.06*10^-2)*x^2

   x = degree of dissociation = 0.0255

percent dissociation = 0.0255*100 = 2.55%

2.

part A

   HBr + NaOH ---> NaBr + H2O

concentration of HBr = 0.1 M

   concentration of NaOH = 0.08 M

so that, excess HBr = 0.1-0.08 = 0.02

pH = -log(H+)

     = -log0.02

     = 1.7

part B

pka of HClO2 = 1.96

pH = pka + log(base/acid)

    = 1.96+log(0.08/(0.1-0.08))

    = 2.562

part C

   pka of benzoic acid (C6H5COOH) = 4.19

pH = pka + log(base/acid)

    = 4.19+log(0.08/(0.1-0.08))

    = 4.8