What is the pH when 0.600 moles of potassium benzoate (KC7H5O2) has been added to 1.50 of water?

1 Answer
Sep 9, 2016

The pH is 8.90.

Explanation:

Our first task is to calculate the concentration of the potassium benzoate,

Molarity=moleslitres=0.600 mol1.50 L=0.400 mol/L

The reaction is

C7H5O-2+H2OHC7H5O2+OH-

For benzoic acid, Ka=6.25×10-5.

For the benzoate ion, as above, Kb=KwKa=1.00×10-146.25×10-5=1.60×10-10

Now, we can set up an ICE table to calculate the concentrations.

mmmmmmC7H5O-2+H2OHC7H5O2+OH-
I/mol⋅L-1:ml0.400mmmmmmmmm0mmmll0
C/mol⋅L-1:mm-xmmmmmmmmll+xmm+x
E/mol⋅L-1:ll0.400 -xmmmmmmmmlxmmmlx

Kb=[HC7H5O2][OH-][C7H5O-2]=x×x0.400x=x20.400x=1.60×10-10

0.4001.60×10-10=2.50×109

x0.400

The equation reduces to

x20.400=1.60×10-10

x2=0.400×1.60×10-10=6.40×10-11

x=8.00×10-6

[OH-]=8.00×10-6lmol/L

pOH=log[OH-]=log(8.00×10-6)=5.10

pH=14.00 - pOH=14.00 - 5.10=8.90