Answer to Question #151027 in Chemistry for Emm

Question

A solution containing 2.00 g of Sr(OH)2 in 250 mL of water was prepared. The PH expected is?

Answer 1

Solution:

Calculate the molarity of a solution:

Molarity = Moles of solute / L of solution

Moles of solute = Mass of solute / Molar mass of solute

or

Molarity = Mass of solute / (Molar mass of solute × L of solution)

The molar mass of Sr(OH)₂ is 121.63 g/mol.

Hence,

Molarity of a solution = (2.00 g) / (121.63 g/mol × 0.25 L) = 0.06577 mol/L = 0.06577 M

We assume Sr(OH)₂ is a strong base, such that

Sr(OH)₂(aq) → Sr²⁺(aq) + 2OH⁻(aq)

and thus, it gives rise to 0.06577 × 2 = 0.13154 M OH⁻.

As a result,

pOH = −log[OH⁻] = −log(0.13154) = 0.88

At any temperature, pH + pOH = pK_w,

and at 25^∘C, pK_w = 14.

Hence,

pH = 14 − 0.88 = 13.12

pH = 13.12

Answer: pH = 13.12 is expected.