Question
Asked Jul 28, 2019

How much NaOH (in moles) must be added to 1 L of a buffer solution that a 1.8 M in acetic acid and 1.2 M in sodium acetate to result in buffer solution of pH 5.22. Assume volume to remain constant. 

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Step 1

A buffer solution is that solution which resists the change in its pH when small amount of acid or base is added to it. Acetate buffer is composed of weak acid and its salt with strong base in a specific concentration .

Step 2

As per question, the required pH of buffer is 5.22, concentration of acetic acid = 1.8 M, concentration of sodium acetate = 1.2 M

pKa of acetic acid = 1.8 x 10-5, pKa = -log (1.8 x 10-5) = 4.75

According to Henderson Hasselbalch equation for acetate buffer, 

 

[Sodium acetate]
pH 3 pК, + 1og
[Acetic acid]
5.22 4.75log [Sodium acetate]
[Acetic aci d]
0.47= log Sodium acetate]
[Acetic acid]
[Sodium acetate]=10047
[Acetic acid]
[Sodium acetate]_2 95
[Acetic acid]
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[Sodium acetate] pH 3 pК, + 1og [Acetic acid] 5.22 4.75log [Sodium acetate] [Acetic aci d] 0.47= log Sodium acetate] [Acetic acid] [Sodium acetate]=10047 [Acetic acid] [Sodium acetate]_2 95 [Acetic acid]

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Step 3

Let the moles of NaOH to be added = x

Required ratio of [sodium acetate]:[acetic acid] = 2.95

As the NaOH added will react with acetic acid to form sodium acetate, [sodium acetate] will increase while [acetic...

295 Sodium acetate]
[Acetic acid]
1.2+x
2.95 =
[1.8-x
2.95(1.8-x)12+ x
5.3-2.95x 1.2+x
3.95x-4.1
x-1.0
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295 Sodium acetate] [Acetic acid] 1.2+x 2.95 = [1.8-x 2.95(1.8-x)12+ x 5.3-2.95x 1.2+x 3.95x-4.1 x-1.0

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