Question

What volume of 0.818 M KOH solution is required to make 3.79 L of a solution with pH of 11.9?

Express your answer with the appropriate units.

Answer 1

Given:
pH = 11.9

use:
pH = -log [H+]
11.9 = -log [H+]
[H+] = 1.259*10^-12 M

use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(1.259*10^-12)
[OH-] = 7.943*10^-3 M

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 0.818 M
M2 = 7.943*10^-3 M
V2 = 3790 mL

use:
M1*V1 = M2*V2
V1 = (M2 * V2) / M1
V1 = (7.943*10^-3 * 3790)/0.818
V1 = 36.8 mL
Answer: 36.8 mL