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The solubility of CaSO4 in pure water at 0oC is 1.09 gram(s) per liter. The value of the solubility product is g

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Answer: See image attached!

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The value of the solubility of product for CaSO₄ in pure water at 0 °C is 1.1881

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

CaSO₄(aq) → Ca²⁺(aq) + SO₄²⁻(aq)

From the balanced equation above,

1 mole of CaSO₄ contains 1 mole of Ca²⁺ and 1 mole of SO₄²⁻

How to determine the concentration of Ca²⁺ and SO₄²⁻

From the balanced equation above,

1 mole of CaSO₄ contains 1 mole of Ca²⁺ and 1 mole of SO₄²⁻

Therefore,

1.09 g/L of CaSO₄ will also contain

  • 1.09 g/L of Ca²⁺
  • 1.09 g/L of SO₄²⁻

How to determine the solubility of product

  • Concentration of Ca²⁺ = 1.09 g/L
  • Concentration of SO₄²⁻ = 1.09 g/L
  • Solubility of product (Ksp) =?

CaSO₄(aq) → Ca²⁺(aq) + SO₄²⁻(aq)

Ksp = [Ca²⁺] × [SO₄²⁻]

Ksp = 1.09 × 1.09

Ksp = 1.1881

Learn more about solubility of product:

https://brainly.com/question/4530083

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