Question

Your boss asks you to prepare approximately 1 L of a 0.10 M buffer solution of acetic acid (Ka = 1.8x10^(-5)). As a first step, you add 0.10 mol of acetic acid to 900 mL of water. You select 6.0 M NaOH as your base. How many milliliters base would you add?

Edit: The question does not specify a pH for the buffer. Is there any way to solve without pH?

Answer 1

Initially only acid is present in the solution. When you add NaOH, it will react with the acetic acid to produce sodium acetate and water.

Since sodium acetate is a salt, it would dissociate in water to give the conjugate base CH3COO-.

Assume that you add 'x' mL of the NaOH base.

Moles of base added = Molarity*Volume in liters = 6*(x/1000) = 0.006*x moles

So, moles of acid would decrease by this amount, and that of conjugate base would increase by this amount.

So after addition of NaOH we have at equilibrium:

Moles of acetic acid = 0.1-(0.006*x)

Moles of conj. base = 0.006*x

Overall moles of acid and conj. base = 0.1-(0.006*x) + 0.006*x = 0.1

Since final volume has to be 1 L, so the overall molarity of the buffer solution would be 0.1 M irrespective of whatever amount of NaOH we add.

Now, for acetic acid we have:

pKa = 4.75

Since you have to prepare a buffer solution, its pH must be close to its pKa value in order to be an effective buffer.

The buffering capacity is greatest when pH = pKa

So,

Using Henderson Hasselbach equation:

pH = pKa + log([conj. base]/[acid])

When pH = pKa, we have:

[conj. base] = [acid]

So,

0.006*x = 0.1-(0.006*x)

Solving we get:

x = 8.33 mL