What concentrations of acetic acid (pKa=4.76) and acetate would be required to p

ID: 636942 • Letter: W

Question

What concentrations of acetic acid (pKa=4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 4.9 ?

Strategy

Rearrange the Henderson–Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A?]/[HA] .

Use the mole fraction of acetate to calculate the concentration of acetate.

Calculate the concentration of acetic acid.

Step 1: The ratio of base to acid is 1.4.

Step 2: The mole fraction of acetate is 0.58, and the concentration of acetate is 0.087.

Step 3: Calculate the concentration of acetic acid.

That is, there are 1.4 molecules of acetate for each molecule of acetic acid.

ANS)

From above information that

given pKa=4.76, pH =4.9

we know that the

Henderson Hasselbalch equation can be written as

pH= pKa+ log{[A-]/[HA]}

4.9= 4.76 + log [A-]/[HA]

0.14= log[ A-]/[HA]

[A-]/[HA] =10^0.14

[A-]/[HA] =1.4

Addition of one on both sidees give

1+[A-]/[HA] =2.4

[HA]+[A-]/ [HA] =2.4

[HA]/ {[HA-]+[A]} =1/2.4

=0.42

n2=1-n1=1-0.42=0.58

the concentration of acetate=n2*c

[Acetate]=0.58*0.15

[Acetate]=0.087 M

Concencentration of acetic acid

=0.42*0.15= 0.063 M

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