Creating Ph buffers

Question

"Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. How many grams of CH3COONa would it be? (Note: the pKa of CH3COOH = 4.7.) Use the Henderson-Hasselbalch equation to calculate these volumes."

Answer 1

pH=pKa+log([CH₃COO^¯] ÷ [CH₃COOH])

4.7=4.7+log(x)

x=1

use equal volumes

5.7=4.7+log(x)

x=10

[CH₃COO^¯] / [CH₃COOH] = 10/1

if volume CH₃COO^¯=b, and volume CH₃COOH=a, then total volume must equal 10

10=a+b

and volume b must be 10 times greater than a, so b=10a

therefore

10=a+10a

10=11a

a=10/11=0.91

b=10a=9.1mL of CH₃COO^¯

a=0.91=0.91 mL of CH₃COOH

Hope this helps.

Answer 2

The Henderson-Hasselbalch equation is pH = pK_a + log [salt] / [acid].

To make a buffer of pH = 4.7, place this value in the Henderson-Hasselbalch equation and solve for the ratio of salt to acid concentrations. This gives 0 = log [salt] / [acid]. Taking the antilog of both sides yields [salt] / [acid] = 1. This means that the ratio of concentrations has to be 1:1. As the starting concentrations are the same, this means that 5 mL of each solution should be mixed to give the pH of 4.7.

For a pH of 5.7, the Henderson-Hasselbalch equation is 5.7 = 4.7 + log [salt] / [acid] or

1 = log [salt] / [acid]. Taking antilog of both sides gives [salt] / [acid] = 10.

In this case the ratio of the concentrations used should be 10:1. So, if X ml of acid is used, then 10 - X mL of salt solution must be used. As the starting concentrations are the same, 10-X / X =10,

so X = 0.91 mL for the acid and 9.1 mL for the salt.