Question

You need to prepare an acetate buffer of pH 5.53 from a 0.842 M acetic acid solution and a 2.60 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.53? The pKa of acetic acid is 4.76.

Answer 1

Using Henderson Hasselbach equation:

pH = pKa + log(moles of salt/moles of acid)

Putting values:

5.53 = 4.76 + log(moles of salt/moles of acid)

Solving we get:

(moles of salt/moles of acid) = 5.89

Initial moles of acid = 0.975*0.842 = 0.82095

Assume that 'x' L of KOH are used.

Moles of base = x*2.6

So, at equilibrium we have:

Moles of salt = x*2.6

Moles of acid = 0.82095-(x*2.6)

So,

5.89 = (x*2.6)/(0.82095-(x*2.6))

Solving we get:

x = 0.269 L

So volume of KOH solution needed = 269 mL

Hope this helps !