I am not understanding this question, is it ok if I ask for help on this one

Question

Calculate the volume of 0.500 M HCOOH and 0.500 M HCOONa required to prepare 0.100 L of pH 5.00 buffer with a buffer strength of 0.100 M. The p𝐾a of HCOOH is 3.75.

volume of HCOOH =L?

volume of HCOONa=L?

Answer 1

Buffer capacity β is given by

β = 2.303 ([H⁺] + T_HA K_a [H⁺] / (K_a + [H⁺])² + K_w / [H⁺] )

You can start by solving the above equation for T_HA which is the total concentration of the buffer

Here β = 0.1 M is the desired buffer capacity (strength)

[H⁺] = 1 x 10^-5 M (pH = 5)

K_a = 1.77 x 10^-4 M (pK_a = 3.75)

K_w = 10^-14

Solving for T_HA gives T_HA = 0.861 M = [HCOOH] + [HCOONa]

Next step is to find the concentrations of [HCOOH] and [HCOONa]

We will use the Henderson Hasselbalch equation

pH = pKa + log₁₀ ([A^-]/[HA])

Here pH = 5, pK_a = 3.75

Solving for [A^-] / [HA] = [HCOONa] / [HCOOH] we obtain

[HCOONa] / [HCOOH] = 17.8

Now we have a system of two equations with two unknowns

[HCOOH] + [HCOONa] = 0.861 M

[HCOONa] / [HCOOH] = 17.8

Solving this system of equations gives us

[HCOOH] = 0.0458 M

[HCOONa] = 0.815 M

The desired buffer solution has the volume of 0.100 L, hence it should contain 0.00458 mole of HCOOH and 0.0815 mole of HCOONa. That means that you need to measure out

V_HCOOH = 0.00458 mole / 0.500 mole/L = 0.00916 L of 0.500 M HCOOH solution

and

V_HCOONa = 0.0815 mole / 0.500 mole/L = 0.163 L of 0.500 M HCOOH solution

The total amount of solution will be more than 0.100 L, hence you will need to get rid of some of the water to reduce the total volume to 0.100 L. It is nontrivial, because formic acid is volatile, so perhaps you need to evaporate (some of) the water from HCOONa solution before adding the HCOOH solution.

Another useful observation is that the buffer we just designed is so weak because its pH (5) is too far from the pKa (3.75) of the acid. Acetic acid / sodium acetate with pKa of 4.76 would have made a pH 5 buffer with much higher capacity. Formic acid / sodium formate is better suited for making a pH 4 buffer.