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A saturated solution of Al(OH)3 has a molar solubility of2.9x J0-9 Mat a certain temperature. What is the solubility product constant, Ksp, of Al(OH)3 at this temperature

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Eduard22sly

The solubility product constant, Ksp, of the aluminum hydroxide, Al(OH)₃ at the temperature is 1.91×10¯³⁹

What is solubility of product?

The solubility of product (Ksp) is defined as the concentration of products raised to their coefficient coefficients. This is illustrated below:

mA <=> nC + eD

Ksp = [C]^n × [D]^e

With the above information in mind, we can obtain the solubility of the product. This is illustrated below:

Dissociation equation

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

How to determine the concentration of Al³⁺ and OH¯

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

Therefore,

2.9×10¯⁹ M Al(OH)₃ will also contain

  • 1 × 2.9×10¯⁹ = 2.9×10¯⁹ M Al³⁺
  • 3 × 2.9×10¯⁹ = 8.7×10¯⁹ M  OH¯

How to determine the solubility of product

Concentration of Al³⁺ = 2.9×10¯⁹ M

Concentration of OH¯ = 8.7×10¯⁹ M

Solubility product (Ksp) =?

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

Ksp = [Al³⁺] × [OH¯]³

Ksp = 2.9×10¯⁹ × (8.7×10¯⁹)³

Ksp = 1.91×10¯³⁹

Learn more about solubility of product:

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