Question

15.24. Calcium hydroxide, also known as slaked lime, is used in industrial processes in which low concentrations of base are required. Only 0.16 g of Ca(OH)2 dissolves in 100 mL of water at 25°C. What is the concentration of hydroxide ions in 250 mL of a solution containing the maximum amount of dissolved calcium hydroxide?​

Answer 1

The concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:

Mass of Ca(OH)₂ = 0.16 g

Molar mass of Ca(OH)₂ = 40 + 2[16 + 1] = 74 g/mol

Mole of Ca(OH)₂ =?

Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

Mole of Ca(OH)₂ = 0.00216 mole

• Next, we shall determine the molarity of the stock solution of Ca(OH)₂.

Mole of Ca(OH)₂ = 0.00216 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of Ca(OH)₂ =?

Molarity = mole / Volume

Molarity of Ca(OH)₂ = 0.00216 / 0.1

Molarity of Ca(OH)₂ = 0.0216 M

• Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 100 mL

Molarity of stock solution (M₁) = 0.0216 M

Volume of diluted solution (V₂) = 250 mL

Molarity of diluted solution (M₂) =?

M₁V₁ = M₂V₂

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

M₂ = 0.00864 M

Thus, the molarity of the diluted solution is 0.00864 M

• Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:

Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)

From the balanced equation above,

1 mole of Ca(OH)₂ contains 2 moles of OH¯

Therefore,

0.00864 M Ca(OH)₂ will contain =  2 × 0.00864 = 0.01728 M OH¯

Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

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