A 1.2×10−5mol sample of Ca(OH)2 is dissolved in water to make up 250.0 mL of solution. What is the pH of the solution at 25.0∘C? Round your answer to the hundredths place.

Answer 1

pH = -log[H₃O⁺]

Molarity = moles

L

250.0 mL x 1L = 0.25L

1000mL

Molarity of Ca(OH)₂ = 1.2 x 10^-5mol Ca(OH)₂

0.25L

Molarity of Ca(OH)₂ = 4.8 x 10^-5M

Ca(OH)₂ is a strong base. The molarity of a strong base is equal to the concentration of OH^-

Ca(OH)₂ has two OH^-, therefore the [OH^-] = 4.8 x 10^-5M x 2.

[OH^-] = 9.6 x 10^-5M

[OH^-][H₃O⁺] = kw (1.0 x 10^-14)

[H₃O⁺] = kw

[OH^-]

[H₃O⁺] = 1.0 x 10^-14

9.6 x 10^-5

[H₃O⁺]= 1.04 x 10^-10M

pH= -log[1.04 x 10^-10M]

pH= 9.98