Question

Calculate the mass of solid formic acid (MM = 46.0 g/mol - Ka = 1.8 x 10-4 ) and sodium formate (MM = 68 g/mol) needed to make 100 mL a 0.2 M formic acid buffer that is 1:3 (conjugate base/weak acid). Also predict the pH of this buffer.

Answer 1

We need to make 100 mL of 0.2M formic acid.

Also, the ratio of conjugate base (sodium formate) : weak acid (formic acid) is1:3 .

So, concentration of conjugate base is  1/3 rd of formic acid concentration = 0.2 M x 1/3 = 0.067 M

Thus, 100 mL solution will be 0.2 M in formic acid and 0.067M in sodium formate.

Now, Molarity = no. of moles/ volume of solution(in L).

So,no. of moles needed = molarity x volume of solution (in L)

Volume of solution =100mL = 0.1 L

So, no. of moles of formic acid needed = 0.2 M x 0.1 L = 0.02 moles

No. of moles of sodium formate needed = 0.067 M x 0.1 L = 0.0067 moles

Also, mass required = no. of moles x molar mass

So, mass of formic acid needed =0.02 moles x 46.0 g/mol = 0.92 g

Mass of sodium formate needed = 0.0067 moles x 68 g/mol = 0.4556 g

To calculate pH of the solution you need to use the Henderson Hasselbach equation:

[Salt] pH = pka + lograridi

Ka is acid dissociation constant. = 1.8 x 10-4

pKa =-logKa = -log(1.8 x 10-4)

So,pKa = 3.74

Here,salt = sodium formate  and acid = formic acid

[…] denotes concentration in molarity.

So, [salt] = 0.067 M,and [Acid] = 0.2 M

Thus, [salt]/[acid] = 0.067/0.2 = 0.33

([salt]/[acid] = 1/3, as given in the question)

(the salt will dissociate to give the formate anion, which is known as the conjugate base of formic acid)

Thus, putting the values in the equation:

pH = 3.74 + log0.33

or, pH = 3.26