Question

You need to prepare an acetate buffer of pH 5.21 from a 0.635 M acetic acid solution and a 2.86 M KOH solution. If you have 830 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.21? The p?a of acetic acid is 4.76.

Answer 1

Acetic acid, CH₃COOH is a weak acid and it dissociates to give CH₃COO^- and H⁺.

So, K_a dissociation constant is given as follows :

Ka=[CH₃COO^-][H⁺]/[CH₃COOH]

pK_a of acetic acid is given as 4.76. Then, K_a value can be calculated using the following formula

pK_a= - logK_a , then the value of K_a is found to be 1.73 x 10^-5 . Similarly, the concentration of H⁺ ions can be found using the formula pH= -log[H⁺]. The [H⁺] is 6.16 x 10^-6 mo/l

∴[CH3COOH]/[CH3COO−] = 6.16×10^-6//1.73×10^-5 = 0.3560

To get the required pH we need to set the acid : salt ratio to this value

To make up buffer of required pH, we need to set the acid: salt ratio.

We add KOH, which gives :

CH3COOH+OH^-→CH3COO^- + H₂O

The above equation shows that the number of moles of OH^- ions is equal to the number of moles of CH₃COO^- ions produced in solution.

Let V be the volume of OH^- ions needed . The initial concentration of acetic acid is given concentration of acetic acid and given volume of acetic acid in liters.

nCH3COOH initial =0.635×0.830= 0.527

After the alkali has been added the number of moles of acetic acid is to be determined. Because the dissociation is small we will assume that these are a good approximation to the equilibrium moles:

nCH3COOH eqm =(0.527−2.86V)

∴nCH3COOH eqmn/CH3COO^- eqm = 0.527−2.86V / 2.86V= 0.3560

on solving the above equation for V, we get the value of V as 0.1361 Litre

or V= 136.1 ml