Question #54a99

1 Answer
Stefan V.
Feb 4, 2015

LONG ANSWER

Start by determining the concentrations of the two species in the buffer. For acetic acid, you have

C=nV⇒nacetic=C⋅V=0.2 M⋅100⋅10−3L=20⋅10−3moles

This means that the concentration of acetic acid in the buffer will be

Cacetic=naceticVbuffer=20⋅10−3moles(100+100)⋅10_3L=0.1 M

For sodium acetate, the number of moles is

nacetate=C⋅V=0.15 M⋅100⋅10−3L=15⋅10−3moles

As a result ,the concentration of sodium acetate in the buffer will be

Cacetate=nacetateVbuffer=15⋅10−3moles(100+100)⋅10−3L=0.075 M

Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution

pHsolution=pKa+log([CH3COO−][CH3COOH])

pHsolution=4.75+log(0.075 M0.1M)=4.75−0.125=4.63

Now you add the HCl solution. The strong HCl acid will react with the wak sodium acetate base to produce weak acetic acid

HCl(aq)+CH3COONa(aq)→NaCl(aq)+CH3COOH(aq)

(I won't go into the net ionic equation)

The moles of hydrochloric acid added to the solution will be

nHCl=C⋅V=0.2 M⋅20⋅10−3L=4⋅10−3moles

This means that the concentration of HCl in the buffer will be

CHCl=nHClVbuffer=4⋅10−3moles(100+100+20)⋅10−3L=0.018 M

The new concetrations of acetic acid and sodium acetate will be

Cacetate=15⋅10−3moles220⋅10−3L=0.068 M, and

Cacetic=20⋅10−3moles220⋅10−3L=0.091 M

Now, all the hydrochloric acid will be consumed by the above reaction; this means that the concentration of sodium acetate will decrease by how much HCl was consumed, and the concentration of acetic acid will increase by the same amount.

Therefore,

Cacetic-final=Cacetic+CHCl=0.091 M+0.018 M=0.109 M, and

Cacetate-final=Cacetate−CHCl=0.068 M−0.018 M=0.05 M

This means that the solution's pH will now be

pHsol=4.75+log(0.05 M0.109 M)=4.75−0.338=4.41

Notice how little the pH dropped despite the addition of a strong acid.

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