Question

What would be the pH if 10 mL of 0.100 M HCL was added to 1000 mL of pure water?

Answer 1

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.1 M

V1 = 10 mL

V2 = 1000 mL + 10 mL = 1010 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (0.1*10)/1010

M2 = 9.90*10^-4 M

use:

pH = -log [H+]

= -log (9.9*10^-4)

= 3.0044

Answer: 3.00