Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

Sagot :

Answer: See image attached!

Good luck with your future exams!

View image txbboat

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