1. Buffer Solutions & Equilibrium Law David Martin City and Islington College Students Name: …………………………………………………………………..

2. Topic Buffer Solutions and Equilibrium Law Aims  This worksheet provides an introduction to buffer solution theory Level Level 3 Method PowerPoint slides (hand-out) - ALL SLIDES TO PRINT OUT (apart from slide 2) Equipment  Laptop & Projector  Hand-out  Pens Duration >30 minutes

3. Buffer Solutions & Equilibrium Law Examining the effects of adding hydrogen ions to a buffer solution made from 1.0 moldm -3 ethanoic acid and 1.0 moldm -3 sodium ethanoate. 1) Write an equation for the dissociation of ethanoic acid. ____________________________________________________________________ 2) Recall or look up the Ka value of ethanoic acid from your notes ____________________________________________________________________ 3) From the Ka value for ethanoic acid, do you expect that the concentration of free hydrogen ions in solution will be: a) Greater than 1.0 moldm -3 b) About the same as the concentration of the ethanoic acid – about 1.0 moldm -3 c) Very low – much less than 1.0 moldm -3 Answer = 4)Write an equation for the addition of sodium ethanoate to water – the equation does not need to have water in it. ____________________________________________________________________

4. 5) What do you expect the concentration of ethanoate ions to be? a) Very low – about the same as the concentration of hydrogen ions b) About the same as the concentration of the ethanoic acid, 1.0 moldm -3 c) High – twice the concentration of the ethanoic acid Answer = 6) What do you think happens to the H + ions when they are added to a buffer solution? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 7) Which equation is important for the removal of hydrogen ions? ____________________________________________________________________ ____________________________________________________________________ 8) The other equation must be important for the removal of hydroxide OH - ions. Can you explain how the other equation might be able to remove OH - ions? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Buffer Solutions & Equilibrium Law

5. Calculating the pH of a buffer solution Consider a buffer solution made from a weak acid, HA in solution with its sodium salt Na + A -. Complete the equation for the acid dissociation. HA (aq) The write a Ka expression Ka = Which ion is responsible for pH determination? ______________________________ We can rearrange the Ka equation So the pH of a buffer should remain constant as long as the concentrations of HA (un- ionised acid) and the A - (anions) stays constant. This is because Ka remains constant. Buffer Solutions & Equilibrium Law

6. Example Calculate the hydrogen ion concentration of a buffer solution made by dissolving both 0.10 mol of ethanoic acid and 0.1 mol of sodium ethanoate in 1.0 dm 3 of water. Ka (CH 3 COOH) = 1.7 x 10 -5 moldm -3 Write an equation for the dissociation of ethanoic acid – working out the start and equilibrium amounts. CH 3 COOH (aq) CH 3 COO - (aq) +H + (aq) Start Equilibrium The “common ion effect” stops the ethanoic acid splitting up. The sodium ethanoate is fully split up providing huge amounts of CH 3 COO - ions so the CH 3 COOH will not split up to make more. Let the amount of H + ions = x Write an equation for the Ka for ethanoic acid: Ka = Rearrange to make [H + ] the subject: [H + ] = Put in what you know [H + ] = Buffer Solutions & Equilibrium Law

7. Assume x is very small so we can assume: 0.10 + x = 0.10 and 0.1 – x = 0.1 [H + ] =pH = The pH of a buffer can be changed if the amount of acid or salt is altered. Problem Calculate the hydrogen ion concentration of a buffer solution made by dissolving both 0.10 mol of ethanoic acid and 0.2 mol of sodium ethanoate in 1.0 dm 3 of water. Ka (CH 3 COOH) = 1.7 x 10 -5 moldm -3 Buffer Solutions & Equilibrium Law

8. ANSWERS

9. Buffer Solutions & Equilibrium Law

10. 5) What do you expect the concentration of ethanoate ions to be? a) Very low – about the same as the concentration of hydrogen ions b) About the same as the concentration of the ethanoic acid, 1.0 moldm -3 c) High – twice the concentration of the ethanoic acid Answer =B 6) What do you think happens to the H + ions when they are added to a buffer solution? The H+ ions will combine with the ethanoate ions to form ethanoic acid. They are removed from the buffer solution and so the pH of the solution will not change. 7) Which equation is important for the removal of hydrogen ions? The sodium ethanoate equation as it contains a large amount of ethanoate ions. 8) The other equation must be important for the removal of hydroxide OH - ions. Can you explain how the other equation might be able to remove OH - ions? The H + ions react with any OH - ions to form water. The ethanoic acid dissociates to make as many H + ions as required to neutralise the OH - ions. Buffer Solutions & Equilibrium Law

11. Calculating the pH of a buffer solution Consider a buffer solution made from a weak acid, HA in solution with its sodium salt Na + A -. Complete the equation for the acid dissociation. HA (aq) H + + A - The write a Ka expression Ka = [H + ] [A - ] [HA] Which ion is responsible for pH determination? H + We can rearrange the Ka equation[H + ] = Ka [HA] [A - ] So the pH of a buffer should remain constant as long as the concentrations of HA (un- ionised acid) and the A - (anions) stays constant. This is because Ka remains constant. Buffer Solutions & Equilibrium Law

12. Example Calculate the hydrogen ion concentration of a buffer solution made by dissolving both 0.10 mol of ethanoic acid and 0.1 mol of sodium ethanoate in 1.0 dm 3 of water. Ka (CH 3 COOH) = 1.7 x 10 -5 moldm -3 Write an equation for the dissociation of ethanoic acid – working out the start and equilibrium amounts. CH 3 COOH (aq) CH 3 COO - (aq) +H + (aq) Start0.10.10 Equilibrium0.1 – x0.1 + xx acidsalt The “common ion effect” stops the ethanoic acid splitting up. The sodium ethanoate is fully split up providing huge amounts of CH 3 COO - ions so the CH 3 COOH will not split up to make more. Let the amount of H + ions = x Write an equation for the Ka for ethanoic acid: Ka =Ka = [0.1+x][x]= [salt+x][x]_ [0.1-x] [acid-x] Rearrange to make [H + ] the subject: [H + ] =Ka [acid-x] [salt+x] Put in what you know [H + ] =1.7 x 10 -5 x 0.1 – x 0.1 + x Buffer Solutions & Equilibrium Law

13. Assume x is very small so we can assume: 0.10 + x = 0.10 and 0.1 – x = 0.1 [H + ] =Ka [acid]1.7 x 10 -5 x 0.1 = 1.7 x 10 -5 [salt] 0.1 pH = 4.77 The pH of a buffer can be changed if the amount of acid or salt is altered. Problem Calculate the hydrogen ion concentration of a buffer solution made by dissolving both 0.10 mol of ethanoic acid and 0.2 mol of sodium ethanoate in 1.0 dm 3 of water. Ka (CH 3 COOH) = 1.7 x 10 -5 moldm -3 Buffer Solutions & Equilibrium Law Answer [H + ] =Ka [acid]1.7 x 10 -5 x 0.1 = 8.5 x 10 -6 [salt] 0.2 pH = 5.07 A greater concentration of salt pushes the acid equilibrium further to the LHS – so there are less H + ions so the pH of this buffer is less acidic than the buffer in the example.

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