You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 M solut

Question

You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa).

Part A

What is the pH of the benzoic acid solution prior to adding sodium benzoate?

Express your answer using three significant figures.

Part B

How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Express your answer using two significant figures.

Explanation / Answer

part A:

Prior to adding sodium benzoate , procedure is same as of a weak acid.
HA + H2O ----> H3O+ + A-

construct ICE table as follows:

Ka = [H3O+][A-]/[HA] = [x][x]/0.02-x

Ka for benzoic acid =6.5 X 10^-5

therefore 6.5 X 10^-5 =x^2/0.02-x

6.5 X 10^-5 (0.02 -x) =x^2

x2 +6.5X10^-5x - 0.13X10^-5 =0
on solving for this we get x= 1.09 X 10^-3

pH = -log [H3O+] = -log [1.09 X 10^-3] = 2.96



Part B:

moles of benzoic acid = 0.02 mol/L X 1.5(L) = 0.03
moles of sodium benzoate = ?

for this we will use Henderson-hasblanch equation

pH=pKa + log( base/acid)

pKa = -log[Ka] = -log[6.5X10^-5] = 4.18

As we know given pH = 4.00 , put all vales in H-H equation

4 = 4.18 + log(benzoate/benzoic acid)

4 - 4.18 = log(benzoate/0.03)

-0.18 = log(benzoate/0.03)

10^-0.18 = benzoate/0.03

0.66 X0.03 = benzoate

benzoate = 0.0198 M

No. of moles of sodium benzoate = 0.0198 X 1.5 (L) = 0.0297 moles

convert moles into grams as follows:

Molar wt. of sodium benzoate = 144.1g/mol

So, .0297mol of C6H5COONa x (144.1 g/mol)= 4.279

therefore 4.279 g of sodium benzoate is added.

HA + H2O -------> H3O+ + A- INITIAL 0.02M 0 0 CHANGE -x +x +x equilibrium 0.02 -x +x +x

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