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# 2012 topic 18 1 calculations involving acids and bases

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### 2012 topic 18 1 calculations involving acids and bases

1. 1. IB Chemistry Power Points Topic 18 Acids and Baseswww.pedagogics.ca Calculations with Acids and Bases
2. 2. Kw : the ionic product constant of water 2 H2O  H3O+ + OH- H2O  H+ + OH- (simplified) [ H 3O ][OH ] [ H ][OH ] Kw 2 [ H ][OH ] [ H 2O ] [ H 2O ] 14 3  K w 1 10 mol dm (at 25 )
3. 3. Kw : depends on temperature Pure water is always neutral i.e. [H+] = [OH-] T (°C) Kw (mol2 dm-6) pH 0 0.114 x 10-14 7.47 10 0.293 x 10-14 7.27 20 0.681 x 10-14 7.08 25 1.008 x 10-14 7.00 This means the pH value that 30 1.471 x 10-14 6.92 is “neutral” i.e. [H+] = [OH-] changes with temperature 40 2.916 x 10-14 6.77 50 5.476 x 10-14 6.63 100 51.3 x 10-14 6.14 Be aware of this. 99% of the time you can assume the Kw value is 1 x 10-14
4. 4. The Kw value for pure water at 60o C is 9.614 × 10-14. Calculate theconcentration of hydroxide ions and the pH of pure water at this temperature.
5. 5. The Kw value for pure water at 60o C is 9.614 × 10-14. Calculate theconcentration of hydroxide ions and the pH of pure water at this temperature.
6. 6. about pOH pOH = -log [OH-] • just like pH, pOH is a measure of concentration • pH + pOH = pKw • this means that at 25o pH + pOH = 14 • unless otherwise stated (or a temperature different than 25o is given), this relationship applies to all pH problemsFor example: what is the [OH-] for a solution with pH 8.2 measured at 25o C?
7. 7. about pOH pOH = -log [OH-] • just like pH, pOH is a measure of concentration • pH + pOH = pKw • this means that at 25o pH + pOH = 14 • unless otherwise stated (or a temperature different than 25o is given), this relationship applies to all pH problemsFor example: what is the [OH-] for a solution with pH 8.2 measured at 25o C? pH pOH 14 pOH 14 pH pOH 14 8.2 5.8 5.8 6 [OH ] 10 1.58 10
8. 8. Weak acids and weak bases – dissociation reactions with waterUnlike strong acids and bases, weak acids and bases do not dissociate100%. This means the concentration acid or base is NOT the same as theequilibrium concentrations of hydronium or hydroxide ions ethanoic acid: CH3COOH + H2O  CH3COO- + H3O+ [CH3COO ][ H 3O ] 5 Ka 1.738 10 [CH 3COOH ] ammonia: NH3 + H2O  NH4+ + OH- [ NH 4 ][OH ] 5 Kb 1.778 10 [ NH3 ]
9. 9. PracticeCalculate the pH ofa) a 0.75 M solution of ethanoic acid Compare to the pH of a 0.75 M HCl solution
10. 10. PracticeCalculate the pH ofa) a 0.75 M solution of ethanoic acid [CH 3COO ][ H 3O ] 5 Ka 1.738 10 [CH 3COOH ] [ x][ x] 5 1.738 10 [0.75 x] x 0.75 1.738 10 5 0.00361 mol dm 3 pH log(0.00361) 2.44 Compare to the pH of a 0.75 M HCl solution
11. 11. PracticeCalculate the pH ofb) a 0.75 M solution of ammonia
12. 12. PracticeCalculate the pH ofb) a 0.75 M solution of ammonia [ NH 4 ][OH ] 5 Kb 1.778 10 [ NH 3 ] [ x][ x] 5 1.778 10 [0.75 x] 5 3 x 0.75 1.778 10 0.00365 mol dm pH 14 pOH 14 log(0.00365) 11.6
13. 13. Consider the following simplified equation for the reaction of ethanoic acid in water CH3COOH  CH3COO- + H+
14. 14. Consider the following simplified equation for the reaction of ethanoic acid in water CH3COOH  CH3COO- + H+Now write the equation for the reaction of the conjugate base with water:
15. 15. Consider the following simplified equation for the reaction of ethanoic acid in water CH3COOH  CH3COO- + H+Now write the equation for the reaction of the conjugate base with water: CH3COO- + H2O  CH3COOH + OH-
16. 16. Consider the following simplified equation for the reaction of ethanoic acid in water CH3COOH  CH3COO- + H+Now write the equation for the reaction of the conjugate base with water: CH3COO- + H2O  CH3COOH + OH-The equilibrium expression for the reaction of the acid is:The equilibrium expression for the reaction of the conjugate base is:
17. 17. Consider the following simplified equation for the reaction of ethanoic acid in water CH3COOH  CH3COO- + H+Now write the equation for the reaction of the conjugate base with water: CH3COO- + H2O  CH3COOH + OH-The equilibrium expression for the reaction of the acid is: [CH 3COO ][ H ] Ka [CH 3COOH ]The equilibrium expression for the reaction of the conjugate base is: [CH3COOH ][OH ] Kb [CH3COO ]
18. 18. Now write an expression for Ka multiplied by Kb
19. 19. Now write an expression for Ka multiplied by Kb [CH 3COO ][ H ] [CH 3COOH ][OH ] K a Kb [CH 3COOH ] [CH 3COO ]
20. 20. Now write an expression for Ka multiplied by Kb [CH 3COO ][ H ] [CH 3COOH ][OH ] K a Kb [CH 3COOH ] [CH 3COO ]
21. 21. Now write an expression for Ka multiplied by Kb [CH 3COO ][ H ] [CH 3COOH ][OH ] K a Kb [CH 3COOH ] [CH 3COO ] K a Kb [ H ][OH ] K w
22. 22. Consider 2 weak acids, use your data bookletto find the missing values ethanoic acid CH3COOH Ka = 1.738 × 10-5 pKa = methanoic acid HCOOH Ka = 1.778 × 10-4 pKa =now 2 weak bases ammonia NH3 Kb = 1.778 × 10-5 pKb = methylamine CH3NH2 Kb = 4.365 × 10-4 pKb =
23. 23. Consider 2 weak acids, use your data bookletto find the missing values ethanoic acid CH3COOH Ka = 1.738 × 10-5 pKa = 4.76 conclusions? methanoic acid HCOOH Can you relate acid/base Ka = 1.778 × 10-4 pKa = 3.75 strength to pK values?now 2 weak bases Easy math between Ka ammonia NH3 and pKa? Kb = 1.778 × 10-5 pKb = 4.75 methylamine CH3NH2 Kb = 4.365 × 10-4 pKb = 3.36
24. 24. More neat stuff.CH3COO- is the conjugate base of the weak acid CH3COOH(Ka value 1.738 × 10-5).CH3COO- is therefore a weak base with a Kb value of 5.75 ×10-10.What do you notice about the pK values for this acid -conjugate base pair? ethanoic acid CH3COOH Ka = 1.738 × 10-5 pKa =4.76 ethanoate anion CH3COO- Ka = 5.75 × 10-10 pKb = 9.24
25. 25. ethanoic acid CH3COOH Ka = 1.738 × 10-5 pKa =4.76 ethanoate anion CH3COO- Ka = 5.75 × 10-10 pKb = 9.24for acid/conjugate base pairs & base/conjugate acid pairs pKa + pKb = pKw Not to be confused with pH + pOH = pKw