Question

A biochemist prepares two stocks of sodium phosphate buffer by titrating phosphoric acid with sodium hydroxide. She prepares a 0.10 M solution of sodium phosphate buffer at pH 2.15, and she prepares a 0.10 M solution of sodium phosphate buffer at pH 7.20. If she mixes 50. mL of the pH 2.15 buffer with 30. mL of the pH 7.20 buffer, what will be the pH of the resulting solution? Recall that the pKas of phosphoric acid are 2.15, 7.20, and 12.35.

2.15

12.35

3.00

8.05

10.94

2.60

9.78

11.50

7.20

4.68

6.35

7.60

Answer 1

the correct answer to this should be 2.7129.

I am giving here the step by step procedure to obtain this answer.

0.1 M of pH is:

pH= -log H⁺

So log H⁺ =1/pH= 1/ 2.15= 0.465

H⁺ =2.918

now mmol of H+ = H⁺ * V_solution = 2.918 * 50 = 145.8 mmol

Also 0.1 M of 7.2 pH:

log H⁺ = 1/pH= 1/7.2= 0.138

H⁺ = 1.37

mmol of H⁺ = H⁺ * V_solution = 1.37 * 30= 41.22 mmol

NOW resultant pH of mixture solution is:

total mmol= 145.8+ 41.22= 187.02

H⁺ = total mmol/ total volume

=187.02/80

=2.33

now; pH=- log H⁺

= - 2.33

=2.7129