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1 markSolubility Product

CAPE Chemistry Unit 1 · 2012 · Paper 2 · Question 2(b)(ii)

The solubility product, Ksp, at 25 °C for calcium carbonate (CaCO3) was found to be 5.0 × 10⁻⁹ mol² dm⁻⁶.

Write the equation for the dissociation of calcium carbonate.

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Other parts of this question

  1. 2(a)Describe, using FIVE essential steps, an experiment which can be used to determine the solubility product of Ca(OH)2 at room temperature.[5 marks]
  2. 2(b)(i)Define the term 'solubility product'.[1 mark]
  3. 2(b)(iii)Write the solubility constant expression for calcium carbonate.[2 marks]
  4. 2(c)(i)Calculate the solubility of calcium carbonate (Ksp = 5.0 × 10⁻⁹ mol² dm⁻⁶ at 25°C) in pure water.[2 marks]
  5. 2(c)(ii)Calculate the solubility of calcium carbonate (Ksp = 5.0 × 10⁻⁹ mol² dm⁻⁶ at 25°C) in 0.1 mol dm⁻³ Na₂CO₃ solution.[3 marks]
  6. 2(d)What is responsible for the difference between the solubilities in (c) (i) and (c) (ii) above?[1 mark]

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