Question

You need to prepare an acetate buffer of pH 5.83 from a 0.853 M acetic acid solution and a 2.61 M KOH solution. If you have 825 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.83? The pKa of acetic acid is 4.76.

Answer 1

In this case, acetic acid, CH3COOH, is the weak acid and the acetate anion, CH3COO−, is its conjugate base. pH=pKa+log([CH3COO⁻] / [CH3COOH])

Acetic acid has pKa=4.76

We need to make a buffer of pH 5.83, which is higher than the pKa of the acid, which means the buffer will contain a lot more conjugate base than weak acid.

So,

[CH3COO−] > [CH3COOH]

which implies

[CH3COO−] / [CH3COOH] >1

Start with 825 mL of a 0.853 M acetic acid solution. The number of moles of acetic acid present in the initial solution will be equal to

825mL*(1L/10³mL)*(0.853 moles CH₃COOH/1L)

= 0.703 moles of CH3COOH

When we add potassium hydroxide, a strong base, the hydroxide anions delivered to the solution by the salt will react with the acid to form acetate anions.

CH3COOH(aq)+OH−(aq)→CH3COO−(aq)+H2O(l)

Notice that it takes 1 mole of hydroxide anions, i.e. 1 mole of potassium hydroxide, to react with 1 mole of acetic acid in order to produce 1 mole of acetate anions.

Now, let's say that x represents the number of moles of acetic acid and y represents the number of moles of acetate anions present in the target solution.

If we take V to be the volume of potassium hydroxide solution, expressed in milliliters, added to the initial acetic acid solution, we can say that, at equilibrium, the buffer will contain

[CH3COOH]=x moles / {(825+V)*10⁻³L}={(10³*x) / (825+V)} M

[CH3COO⁻]= y moles / {(825+V)*10⁻³L}={(10³*y) / (825+V)} M

[CH3COO−] / [CH3COOH]={(10³*y) / (825+V)M} / {(10³*x) / (825+V)M} =y/x

According to the Henderson - Hasselbalch equation, we will have

5.83 =4.76+log(y/x)

This is equivalent to

log(y/x)= 1.07

10^log(y/x)=10^1.07

So, y/x= 11.74

Let's get back to the balanced chemical equation that describes the neutralization reaction. If we use n as the number of moles of hydroxide anions added to the initial solution, you can say that the buffer will contain

x=0.703−n → the number of moles of acetic acid

y=0+n→ the number of moles of acetate anions

So,

y/x=n/ (0.703−n) = 11.74

Solving this, we get n= 0.647

Therefore, we must add 0.647 moles of hydroxide anions to the initial acetic acid solution.

Using its molarity to figure out the volume that would contain this number of moles of potassium hydroxide

0.647moles KOH * (1 L solution / 2.61moles KOH) =0.247 L solution = 247 millilitres of KOH solution will be needed to make a buffer of pH 5.83