STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 8)

by BerryBerryTeacher

in Berry Reference (Notes)

Ionic product of water, Kw is an equilibrium constant for reactions. Part 8 of Berry Berry Easy’s notes on Ionic Equilibrium for STPM Chemistry Form 6 is fully about ionic product of water. It should be noted that the relationship between Kw and pKw is exactly the same as that between Ka and pKa. Hence, it’ll be advisable to read up Parts 5-7 before starting Part 8 on this post.

(Tips: Most Berry Readers will typically think of constants as being static and non-changeable. However, it must be noted that the ionic product of water, Kw varies with temperature. As such, do take note of the temperatures of the example given. Make a mental note and ask yourself why is it that most examples are given for 25˚C. By understanding this point, you’ll be better of understanding the concept of constants and it can vary with temperature or other conditions in some cases.)

STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 8 )

Natural Universal Indicator_Purple cabbage

Natural Universal Indicator_Purple cabbage

Ionic product, Kw of water

H2O(l) <—-> H+(aq) + OH-(aq)

  • Kc = ([H+] [OH- ]) / [H2O]
  • Kw = [H+] [OH- ] where Kw = Kc [H2O]
  • In pure water at 25˚C, [H+] = [OH- ] = 10-7 mol dm-3
  • Ka (acid) x Kb (conjugate base) = 10-14 at 25˚C
  • pKa + pKb = 14.0 = pKw

pH scale

  • pH = -lg [H3O+] or pH = -lg [H+]
  • pOH = -lg [OH- ]
  • pH + pOH = 14.0
  • pH = 14.0 + lg [OH- ]
  • pH = 14.0 – pOH

pH values of pure water [H3O+] = 1.0 x 10-7 mol dm-3

  • pH = -lg [H3O+] or pH = -lg [H+]
  • pH = -lg (1.0 x 10-7)
  • pH = – (-7.0)
  • pH = 7.0

pH values of strong acids of 1.0 mol dm-3 (example: HCl)

  • pH = -lg [H3O+] or pH = -lg [H+]
  • pH = -lg 1.0 = 0.0

pH values of strong bases of 1.0 mol dm-3

  • pOH = -lg [OH- ]
  • pOH = -lg 1.0 = 0.0
  • pH = 14.0 – pOH
  • pH = 14.0 – 0.0 = 14.0

pH values of weak acids of 0.15 mol dm-3 (Ka for the acid at 25˚C is 5.80 x 10-10 mol dm-3) (example: HCN)

  • for weak acid, [H+] = √(Ka x c) or [H3O+] = √(Ka x c)
  • for weak acid [H3O+] = √(5.80 x 10-10 x 0.15)
  • [H3O+] = 9.33 x 10-6 mol dm-3
  • pH = – lg (9.33 x 10-6)
  • pH = 5.03

pH values of weak bases of 0.15 mol dm-3 (Kb for the base at 25˚C is 7.30 x 10-10 mol dm-3)

  • for weak base, [OH- ] = √(Kb x c)
  • for weak base [OH- ] = √(7.30 x 10-10 x 0.15)
  • [OH-] = 1.05 x 10-5 mol dm-3
  • pOH = – lg (1.05 x 10-5)
  • pOH = 4.98
  • pH = 14.00 – 4.98
  • pH = 9.02

Bonus: One final question to ask yourself as added bonus. Does the pH of water remain the same at 7.0 for all temperatures? For those that answer yes, you are wrong. At high temperature such as 100˚C, the pH value is 6.14. So water is acidic at 100˚C? If you answer yes to that question, you are wrong again. It is just that the neutral point has shift to 6.14 for that temperature. While this might not come out in your exams, but it is always good to know to have a better understanding of the concept.

Stay tuned to Part 9 of Ionic Equilibrium note for STPM Chemistry Form 6 students from Berry Berry Easy. We only bring you the best and most concise notes for your reading pleasure.

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