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Acid base equilibria

Acid base equilibria. Starter.

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Acid base equilibria

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  1. Acid base equilibria

  2. Starter 44.0g of ethyl ethanoate was mixed with 36.0g of water containing HCl as a catalyst and allowed to reach equilibrium over several days. The equilibrium mixture was then made up to 250cm3 with pure water. A 25.0 cm3 sample of diluted mixture was titrated with 1.00mol dm-3 NaOH. After allowing for the acid catalyst present, the ethanoic acid in the equilibrium mixture was found to require 29.5 cm3 of NaOH for neutralisation. Find the Kc for the reaction: CH3COOC2H5(l)+ H2O(l) CH3COOH(l) + C2H5OH(l) The equation of for the neutralisation reaction is: CH3COOH + NaOH  CH3COONa + H2O

  3. Starter 44.0g of ethyl ethanoate was mixed with 36.0g of water containing HCl as a catalyst and allowed to reach equilibrium over several days. The equilibrium mixture was then made up to 250cm3 with pure water. A 25.0 cm3 sample of diluted mixture was titrated with 1.00mol dm-3 NaOH. After allowing for the acid catalyst present, the ethanoic acid in the equilibrium mixture was found to require 29.5 cm3 of NaOH for neutralisation. Find the Kc for the reaction: CH3COOC2H5(l)+ H2O(l) CH3COOH(l) + C2H5OH(l) The equation of for the neutralisation reaction is: CH3COOH + NaOH  CH3COONa + H2O = 0.249 (no units)

  4. Learning Objectives • Define pH • Calculate pH and H+ ion concentration for strong acids • Define the ionic product for water • Calculate the pH and the OH- ion concentration for strong bases

  5. Defining pH pH = -log10[H+] 100 is 102 the logarithm to base 10 of 100 is 2 1000 is 103 the logarithm to base 10 of 1000 is 3 2 is 100.3010 the logarithm to base 10 of 2 is 0.3010 The logarithm to base 10 of a number is the power you have to raise 10 to in order to equal that number

  6. Using your calculator • To find log102: [2] [log][=] [0.301029995] OR [log] [2][=] [0.301029995] To unlog use the 10x often found above the log button and using the shift key.

  7. Calculating pH from hydrogen ion concentration What is the pH if [H+] = 0.1 mol dm-3 Find log10(0.1) = -1 NB: pH = -log10[H+] “-(-1) is 1” pH = 1

  8. Calculating hydrogen ion concentration from pH NB: hydrogen ion concentrations will almost always be less than 1 mol dm-3 pH is 2.80 what is the hydrogen ion concentration? pH = -log10[H+] Rearranging this: log10[H+] = -pH log10[H+] = -2.80 [shift][log][-][2.8][=][1.58 x 10-3] = 1.58 x 10-3 mol dm-3

  9. Questions Convert the following hydrogen ion concentrations (all in mol dm-3) into pHs • 0.0100 • 0.0250 • 3.00 x 10-4 • 1.00 x 10-7 • 7.50 x 10-10

  10. Convert the following pHs into hydrogen ion concentrations in mol dm-3 • 3.42 • 1.20 • 5.65 • 8.40 • 13.0

  11. The pH of strong acids • What is the pH of 0.1 mol dm-3 HCl? • Because HCl is a strong acid every 1 mole of HCl is entirely split up into 1 mole of H+(aq) and 1 mole of Cl-(aq). The concentration of hydrogen ions is therefore exactly the same as the concentration of the acid • [H+] = 0.1 • pH = log10[H+] • pH = -log10(0.1) • pH = 1

  12. What is the pH of 0.00100mol dm-3 H2SO4? • H2SO4 2H+(aq) + SO42-(aq) • [H+] = 2 x 0.0100 • pH = – log10(0.0200) • pH = 1.70

  13. Finding the concentration of a strong acid from its pH Simply the reverse the process • Convert pH into [H+] • Use [H+] to find the concentration of the acid

  14. What is the concentration of HCl whose pH is 1.60? • “Unlogging” the pH gives [H+] = 0.0251 mol dm-3 • If you got an answer of 39.8107 you forgot to make the pH ‘-’ before you unlogged it. • As HCl is a monoprotic acid the concentration of the acid is the same as the concentration of the H+ ions.

  15. What is the concentration of H2SO4, if its pH is 1.00? • Sulfuric acid is diprotic. • Unlogging th pH gives [H+] = 0.1 mol dm-3 • Now stop and think: • H2SO4(aq) 2H+(aq) + SO42-(aq) • Each mole af acid gives 2 moles of H+(aq). There are only half the number of moles of acid as of H+ ions. • The concentration of the acid is therefore 0.0500mol dm-3.

  16. Calculate the pHs of the following strong acids • 0.0300 mol dm-3 HCl • 0.00500 mol dm-3 H2SO4 • 0.120 mol dm-3 HNO3

  17. Calculate the pHs of the following strong acids • 0.0300 mol dm-3 HCl 1.52 • 0.00500 mol dm-3 H2SO4 2.00 • 0.120 mol dm-3 HNO30.92

  18. Calculate the concentrations of the following strong acids from their pHs • HCl of pH 0.70 • H2SO4 of pH 1.5 • HNO3 of pH 2.0

  19. Calculate the concentrations of the following strong acids from their pHs • HCl of pH 0.70 0.2 mol dm-3 • H2SO4 of pH 1.5 0.0016 mol dm-3 • HNO3 of pH 2.0 0.010 mol dm-3

  20. Ionic product of water • Whenever liquid water is present, this equilibrium is established • H2O(l) H+(aq)+ OH-(aq) • Just like any other equilibrium, you can write an expression for the equilibrium constant Kc. You might expect it to look like this: Kc = [H+] [OH-] [H2O]

  21. But it doesn’t! such a tiny amount of water ionises that the concentration of the water is effectively constant. You need to go to 9 significant figures before you noticed any change in the concentration of the wtare following ionisation!

  22. Ionic product of water Kw = [H+][OH-] Kw is usually 1.00 x 10-14 The units are [concentration]2

  23. Calculating the pH of pure water • pH varies with temperature, • The ionisation process is endothermic • If the water is pure the concentrations of [H+] and [OH-] will be equal • H2O(l) H+(aq) + OH-(aq)

  24. If Kw is 1.00 x 10-14 mol2dm-6 (at 24oC) • Kw = [H+][OH-] • And [OH-] = [H+] because the water is pure • So Kw = [H+]2 • [H+]2 = 1.00x 10-14 • [H+] = 1.00 x 10-7 • pH = -log10 [H+] • pH = 7.00

  25. If Kw is 5.3 x 10-13 mol2dm-6 (at 100oC) • pH = 6.14 • The neutral point has moved from the more familiar value.

  26. Calculate the pH of pure water at: • 15oC Kw = 4.52 x 10-15 mol2dm-6 • 50oC Kw = 5.48 x 10-14 mol2dm-6

  27. Calculate the pH of pure water at: • 15oC Kw = 4.52 x 10-15 mol2dm-67.17 • 50oC Kw = 5.48 x 10-14 mol2dm-66.63

  28. The pH of strong bases • When most bases dissolve in water, the solution consist of free hydrated metal ions and hydroxide ions • E.g NaOH(s) + aq  Na+(aq) + OH-(aq) • Ca(OH)2(s) + aq  Ca2+(aq) + 2OH-(aq) • Typical pHs are 13 or 14 • This means [H+] is 10-13 or 10-14

  29. Calculating the pH of a strong base • Use the concentration of the base to find [OH-] • Use Kw to find [H+] • Convert [H+] into pH • In the next examples I will take Kw to be 1.00 x 10-14 mol2dm-6

  30. What is the pH of 0.10mol dm-3 NaOH? Each mole of NaOH gives 1 mole of OH- in solution, so the concentration of OH- is also 0.1 mol dm-3 Now use Kw and substitute in the value for [OH-] [H+][OH-] = 1.00 x 10-14 [H+] x 0.1 = 1.00 x 10-14 [H+] = 1.00 x 10-14/0.10 [H+]= 1.0 x 10-13 Convert [H+] into pH = 13

  31. Finding the concentration of a strong base from its pH Again you reverse the process • Convert pH into [H+] • Use Kw to find [OH-] • Use the [OH-] to find the concentration of the base.

  32. What is the concentration of KOH solution if its pH is 12.8? • “Unlog” the pH to give the [H+] pH =-log10[H+] 12.8 = -log10[H+] Log10[H+] = -12.8 [H+] = 1.585 x 10-13

  33. Now use Kw to find [OH-] • [H+][OH-] = 1.00 x 10-14 • 1.585 x 10-13 x [OH-] = 1.00 x 10-14 • [OH-] = 1.00 x 10-14/1.585 x 10-13 • [OH-] = 0.0631 mol dm-3 • Because each mole of KOH produces 1 mole of OH-, the concentration of KOH = 0.0631 mol dm-3

  34. What is the concentration of Ba(OH)2 if its pH is 12.0? “Unlogging” the pH gives [H+] = 1.00 x 10-12 Now use Kw to find [OH-]: [H+][OH-] = 1.00 x 10-14 1.00 x 10-12 x [OH-] = 1.00 x 10-14 [OH-] = 1.00 x 10-14/ 1.00 x 10 -12 [OH-] = 0.0100 mol dm-3 Each mole of Ba(OH)2 gives 2 moles of OH- this means the concentration will only be half = 0.00500mol dm-3

  35. Calculate the pHs of the following strong bases • 0.250 mol dm-3 NaOH • 0.100 mol dm-3 Ba(OH)2 • 0.00500 mol dm-3 KOH

  36. Calculate the pHs of the following strong bases • 0.250 mol dm-3 NaOH 13.4 • 0.100 mol dm-3 Ba(OH)213.3 • 0.00500 mol dm-3 KOH 11.7

  37. Calculate the concentrations of the following strong bases from their pHs • NaOH of pH 13.2 • Sr(OH)2 of pH 11.3

  38. Calculate the concentrations of the following strong bases from their pHs • NaOH of pH 13.2 0.158 mol dm-3 • Sr(OH)2 of pH 11.3 1.00 x 10-3 mol dm-3

  39. Summarise pH [OH-] log10 10x[H+]strong acids diprotic acidsmonoprotic acidsKw Strong bases

  40. Apply to exam questions

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