Question

A 1L buffer solution is 0.150M in HC7H5O2 and 0.250M in LiC7H5O2. Calculate the pH of the soln after the additon of 100mL of 1.00M HCl. The Ka HC7H5O2 = 6.5x10^-5

Answer 1

C=n/V⇒n=C⋅V

n benzoic acid=0.150 M⋅1.00 L=0.150 moles

n benzoate=0.250 M⋅1.00 L=0.250 moles

he number of moles of hydrochloric acid add to the buffer is equal to

nHCl=1.00 M⋅100⋅10^−3L=0.100 moles

The balanced chemical equation for this reaction looks like this

C6H5COO−(aq)+HCl(aq)→C6H5COOH(aq)+Cl−(aq)
I..........0.250......................0.100...................0.150
C.......(-0.100)....................(-0.100)................(+0.100)
F.........0.150..........................0.........................0.250

0.250 moles of benzoate ions and 0.150 moles of benzoic acid, and ended up with 0.150 moles of benzoate ions and 0.250 moles of benzoic acid.

All the hydrochloric acid was consumed by the reaction.

The total volume of the solution will be

Vtotal=Vinitial+VHCl

Vtotal=1.00 L+100⋅10−3L=1.10 L

The new concentrations of the benzoic acid and of the benzoate ions will be

[C6H5COOH]=0.250 moles /1.10 L=0.2273 M

[C6H5COO−]=0.150 moles/1.10 L=0.1364 M

The pKa of the acid is

pKa=−log(Ka)=−log(6.5⋅10^−5)=4.19

Henderson-Hasselbalch equation

pH of the solution

pHsol=pKa+log([C6H5COO−] /[C6H5COOH])

pHsol=4.19+log(0.1364M /0.2273M)=3.97