What is the pH of a solution that is formed by diluiting 450 mL of 0.045 molar NaOH with 300 mL of water?

Question

Assume volumes of liquids are additive.

Answer 1

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Calculate the molarity of the final solution:

M1V1 = M2V2

M1* 750 = 0.045*450

M1 = 20.25/750

M1 = 0.027M

[OH-] = 0.027M

Equation:

[H+][OH-] = 10^-14

[H+] = 10^-14/[OH-]

[H+] = 10^-14/ 0.027

[H+] = 3.70*10^-13

pH = -log [H+]

pH = -log (3.70*10^-13)

pH = 12.43