• Skip to primary navigation
  • Skip to main content
A-Level H2 Chemistry Tuition by 10 Year Series Author

A-Level H2 Chemistry Tuition by 10 Year Series Author

Advanced Chemistry Made Easy

  • A Level H2 Chemistry Tuition
  • Success Stories
  • Books
    • A-Level H2 Chemistry Topical Edition Past Year Series
    • A-Level H2 Chemistry Yearly Edition Past Year Series
  • Videos
  • About Me
  • Contact

Chemistry Questions

Ionic Equilibrium: Question on Calculation of Solubility Product, Ksp

December 16, 2010 By Sean Chua 12 Comments


(Bags of Calcium Hydroxide Powders ready for shipment. Photo credit A. Myers)

Calcium hydroxide, traditionally called slaked lime, is an inorganic compound with the chemical formula Ca(OH)2. It exists as a colourless crystal or white powder and is obtained when calcium oxide (called lime or quicklime) is mixed, or “slaked” with water. It is of low toxicity and enjoys many applications.

Let’s check out a typical exam-based question on Solubility Product under the topic of Equilibria, that has to do with calcium hydroxide.

Question:

(a) Write an expression for the solubility product, Ksp, of calcium hydroxide, Ca(OH)2.

(b) In a titration experiment in the school chemistry laboratory, 20.0 cm3 of an aqueous calcium hydroxide solution is completely neutralised by 18.2 cm3 of hydrochloric acid solution with a molar concentration of 0.050 moldm-3.

(i) Determine the hydroxide ion concentration, and hence the pH of the aqueous calcium hydroxide solution.

(ii) Calculate a value of the solubility product of calcium hydroxide, indicating its units.

(c) State one use of calcium hydroxide which depends on its solubility in water.

Suggested Answers & Comments:

Try out the questions before you click the link below for suggested answers and comments:

(a) Dissociation Equation: Ca(OH)2(aq) ↔ Ca2+(aq) + 2OH–(aq)

Solubility constant, Ksp = [Ca2+][OH–]2

(b)(i) The ionic equation for neutralisation is H+(aq) + OH–(aq) → H2O(l)

Mole of H+ = 0.050 x (18.2/1000) = 9.10 x 10-4 mol

Since 1 mole of H+ reacts with 1 mole of OH–, thus:

Mole of OH– = 9.10 x 10-4 mol

Therefore, [OH–] = (9.10 x 10-4 mol) / (20.0/1000) = 0.0455 moldm-3

pOH = -log10[OH–] = -log10(0.0455) = 1.34

pH + pOH = 14

pH = 14 – pOH = 14 – 1.34 = 12.66 ≈ 12.7 (3 s.f.)

(b)(ii) Dissociation Equation: Ca(OH)2(aq) ↔ Ca2+(aq) + 2OH–(aq)

[Ca2+] = 1/2[OH–] = 0.0455/2 moldm-3

Therefore, Ksp = (0.0455/2)(0.0455)2 mol3dm-9 = 4.71 x 10-5 mol3dm-9

Common Errors on Units:

Many students tend to make mistakes in determining the units of the solubility product. Simple multiplication and being more cautious will help you.

(moldm-3) x (moldm-3)2 = (moldm-3) x (mol2dm-6) = mol3dm-9

(c) Calcium hydroxide is commonly used in agriculture to reduce the acidity of soils i.e. ‘liming the soil’.

Hope you find the above discussions useful. I would love to hear from you. Leave me a comment below.

PS: If you find this blogpost useful to someone your know, feel free to send to them. 😀

  • « Go to Previous Page
  • Page 1
  • Interim pages omitted …
  • Page 3
  • Page 4
  • Page 5
  • Page 6
  • Page 7
  • Interim pages omitted …
  • Page 18
  • Go to Next Page »

Copyright © 2010–2026 · ALevelH2Chemistry.com · Talk to us at 88290998