Question

You are to prepare a pH 3.50 buffer and you have 0.10M solution.

HCOOH           Ka= 1.8x10^-4

CH₃COOH      Ka= 1.8x10^-5

HCOONa

CH₃COONa

How much of each solution would you need to prepare 500.0 mL of the buffer at the required pH?

Answer 1

For formic acid

$K_a= 1.8 \times 10^{-4}$

$pK_a=-log K_a = -log 1.8 \times 10^{-4} =3.74$

For acetic acid

$K_a= 1.8 \times 10^{-5}$

$pK_a=-log K_a = -log 1.8 \times 10^{-5} =4.74$

The pH of buffer solution to be prepared is 3.50. It is close to pKa of formic acid which is 3.74. Hence, we will use formic acid.

$\\ \: \\ \: pH=pK_a + log \dfrac {[HCOONa]}{[HCOOH]} \\ \: \\ \: 3.50=3.74 + log \dfrac {[HCOONa]}{[HCOOH]} \\ \: \\ \: log \dfrac {[HCOONa]}{[HCOOH]} = -0.24 \\ \: \\ \: \dfrac {[HCOONa]}{[HCOOH]}=0.569$

Let us assume that we mix x mL of 0.10 M HCOOH to 500-x mL of 0.10 M HCOONa

$\\ \: \\ \: \dfrac {[HCOONa]}{[HCOOH]}=0.569 \\ \: \\ \: \dfrac {500-x}{x}=0.569 \\ \: \\ \: 500-x=0.569x \\ \: \\ \: 500 = 1.569x \\ \: \\ \: x = 319 \\ \: \\ \: 500-x = 500-319= 181$

Thus, we mix 319 mL of 0.10 M HCOOH to 181 0.10 M HCOONa