Question

What volume of 0.877 M KOH solution is required to make 3.53 L of a solution with pH of 12.0?

Answer 1

pH = -log [H3O+]

12 = -log [H3O+]

[H3O+] = 1.0*10^-12 M

use:

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(1.0*10^-12)

[OH-] = 1.0*10^-2 M

[KOH] = [OH-]

= 1.0*10^-2 M

= 0.01 M

This is final concentration of KOH

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given

M1 = 0.877 M

M2 = 0.01 M

V2 = 3.53 L

Use:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (0.01*3.53)/0.877

V1 = 0.0403 L

V1 = 40.3 mL

Answer: 40.3 mL