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Calculations Involving Acids and Bases IB Chemistry Power Points Topic 18 Acids and Bases www.pedagogics.ca.

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Presentation on theme: "Calculations Involving Acids and Bases IB Chemistry Power Points Topic 18 Acids and Bases www.pedagogics.ca."— Presentation transcript:

1 Calculations Involving Acids and Bases IB Chemistry Power Points Topic 18 Acids and Bases www.pedagogics.ca

2 K w : the ionic product constant of water 2 H 2 O  H 3 O + + OH - H 2 O  H + + OH - (simplified)

3 K w : depends on temperature T (°C)K w (mol 2 dm -6 )pH 00.114 x 10 -14 7.47 100.293 x 10 -14 7.27 200.681 x 10 -14 7.08 251.008 x 10 -14 7.00 301.471 x 10 -14 6.92 402.916 x 10 -14 6.77 505.476 x 10 -14 6.63 10051.3 x 10 -14 6.14 Pure water is always neutral i.e. [H + ] = [OH - ] This means the pH value that is “neutral” i.e. [H + ] = [OH - ] changes with temperature

4 The K w value for pure water at 60 o C is 9.614 × 10 -14. Calculate the concentration of hydroxide ions and the pH of pure water at this temperature.

5 about pOH pOH = -log [OH - ] just like pH, pOH is a measure of concentration pH + pOH = pK w  this means that at 25 o pH + pOH = 14  unless otherwise stated (or a temperature different than 25 o is given), this relationship applies to all pH problems For example: what is the [OH - ] for a solution with pH 8.2 measured at 25 o C?

6 Weak acids and weak bases – reactions with water summary ethanoic acid:CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + ammonia:NH 3 + H 2 O  NH 4 + + OH -

7 Practice Calculate the pH of a) a 0.75 M solution of ethanoic acid b)a 0.75 M solution of ammonia

8 Consider the following simplified equation for the reaction of ethanoic acid in water CH 3 COOH  CH 3 COO - + H + Now write the equation for the reaction of the conjugate base with water: CH 3 COO - + H 2 O  CH 3 COOH + OH - The equilibrium expression for the reaction of the acid is: The equilibrium expression for the reaction of the conjugate base is:

9 Now write an expression for K a x K b

10 ethanoic acid CH 3 COOH K a = 1.738 × 10 -5 pK a = methanoic acid HCOOH K a = 1.778 × 10 -4 pK a = Consider 2 weak acids and 2 weak bases ammonia NH 3 K b = 1.778 × 10 -5 pK b = methylamine CH 3 NH 2 K b = 4.365 × 10 -4 pK b = 4.76 3.75 4.75 3.36 conclusions?

11 More neat stuff. CH 3 COO - is a weak base with a K b value of 5.75 × 10 -10. CH 3 COO - is the conjugate base of the weak acid CH 3 COOH (K a value 1.738 × 10 -5 ). What do you notice about the pK values of this acid - conjugate base pair? for acid/conjugate base pairs & conjugate acid/base pairs pK a + pK b = pK w in addition, for a solution pH + pOH = pK w

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