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Starter 1) Write the expression and value of Kw [H+][OH-]kw 1x 10 -2 1x 10 -12 1.00 x 10 -14 1x 10 -2 1.00 x 10 -14 1x 10 -4 1.00 x 10 -14 1x 10 -5 1.00.

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Presentation on theme: "Starter 1) Write the expression and value of Kw [H+][OH-]kw 1x 10 -2 1x 10 -12 1.00 x 10 -14 1x 10 -2 1.00 x 10 -14 1x 10 -4 1.00 x 10 -14 1x 10 -5 1.00."— Presentation transcript:

1 Starter 1) Write the expression and value of Kw [H+][OH-]kw 1x 10 -2 1x 10 -12 1.00 x 10 -14 1x 10 -2 1.00 x 10 -14 1x 10 -4 1.00 x 10 -14 1x 10 -5 1.00 x 10 -14 1x 10 -7 1.00 x 10 -14

2 Buffers Describe what is meant by a buffer solution. State that a buffer solution can be made from a weak acid and a salt of the weak acid. Explain the role of the conjugate acid–base pair in an acid buffer solution. Calculate the pH of a buffer solution from the K a value of a weak acid and the equilibrium concentrations of the conjugate acid–base pair.

3 A buffer solution is a mixture that minimises pH changes on addition of small amounts of acids and bases

4 Making a buffer solution from a weak acid and a salt of the weak acid

5 Shifting the buffer equilibrium

6 CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq)

7 Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol dm -3 and [A¯] of 0.1 mol dm -3. K a = [H + (aq)] [A¯(aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A¯(aq)]

8 Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol dm -3 and [A¯] of 0.1 mol dm -3. K a = [H + (aq)] [A¯(aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A¯(aq)] from information given[A¯] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3

9 Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol dm -3 and [A¯] of 0.1 mol dm -3. K a = [H + (aq)] [A¯(aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A¯(aq)] from information given[A¯] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3 If the K a of the weak acid HA is 2 x 10 -4 mol dm -3. [H + (aq)]= 0.1 x 2 x 10 -4 = 2 x 10 -4 mol dm -3 0.1

10 Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HA] is 0.1 mol dm -3 and [A¯] of 0.1 mol dm -3. K a = [H + (aq)] [A¯(aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A¯(aq)] from information given[A¯] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3 If the K a of the weak acid HA is 2 x 10 -4 mol dm -3. [H + (aq)]= 0.1 x 2 x 10 -4 = 2 x 10 -4 mol dm -3 0.1 pH= - log 10 [H + (aq)] = 3.699

11 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3.

12 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)]

13 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)]

14 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore [HX] = 0.05 mol dm -3 and [X¯] = 0.10 mol dm -3

15 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10 -5 mol dm -3. K a = [H + (aq)] [X¯(aq)] [HX(aq)] re-arrange[H + (aq)] = [HX(aq)] K a [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore [HX] = 0.05 mol dm -3 and [X¯] = 0.10 mol dm -3 Substituting[H + (aq)] = 0.05 x 4 x 10 -5 = 2 x 10 -5 mol dm -3 0.1 pH= - log 10 [H + (aq)] = 4.699


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