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Starter question.

Starter question. Consider propanoic acid (CH 3 CH 2 COOH) reacting with water….. 1. Write a symbol equation for the equilibrium that is established. 2. Identify the conjugate base and acid on the right hand side of the equation. 3. Write a K c expression for the equilibrium. Starter answers.

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Starter question.

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  1. Starter question. • Consider propanoic acid (CH3CH2COOH) reacting with water….. • 1. Write a symbol equation for the equilibrium that is established. • 2. Identify the conjugate base and acid on the right hand side of the equation. • 3. Write a Kc expression for the equilibrium.

  2. Starter answers • CH3CH2COOH (aq) + H20(l) ↔ H3O+(aq) + CH3CH2COO-(aq) • Conjugate base….CH3CH2COO- • Conjugate acid H3O+ • Ka = [H+(aq)][CH3CH2COO-(aq)] • [CH3CH2COOH (aq)]

  3. The pH scale

  4. The pH scale • Devised by a Danish chemist called Soren Sorensen. • It indicates how much acid or alkali is present in a solution. • The ‘p’ in pH stands for ‘potens’, which is latin for power. • pH = -log[H+ (aq)]

  5. pHScale

  6. Calculating the pH of a strong acid.

  7. Calculating the pH of a strong acid • The pH of strong acids are relatively simple to calculate. • If we assume that the acid molecules fully dissociate then [H+] = [HA] • Therefore pH = -log [HA] • E.g. 1 mol HCl, pH = –log[1] = 0 • E.g. 0.1 mol HCl, pH = -log[0.1] = 1

  8. Calculating the pH of a weak acid.

  9. Calculating the pH of a weak acid. An example. • What is the pH of 0.1moldm-3ethanoic acid, which has a Ka of….1.74X10-5moldm-3?) • You need to… • 1. Write out the equilibrium expression (Ka) • 2. convert hydrogen concentration [H+] to pH

  10. Step 1 the Ka expression. Equilibrium equation. CH3COOH (aq) ↔ H+ (aq) +CH3COO-(aq) So the general Ka expression will be…… Ka = [CH3COO-(aq)][H+(aq)] [CH3COOH(aq)] But we need to consider when full equilibrium has been reached and take account of our initial concentration of acid.

  11. Step 2 the equilibrium position. CH3COOH ↔ H+ +CH3COO- CH3COOH For every ethanoic acid molecule that dissociates an ethanoate ion and a H+ also forms. H+ CH3COO-

  12. Writing the new Ka expression at equilibrium. Initial concentration minus the dissociation concentration of the ions. • Ka =[CH3COO-(aq)]eq[H+ (aq)]eq • 0.100 – [CH3COO-(aq)]eq • As [CH3COO-]= [H+] at equilibrium we can write… • Ka = [H+]2 • 0.100 – [H+] Because the disassociation of the weak acid is so small we assume 0.100 – [H+] = 0.100

  13. Plugging the numbers in.. Substituting in the values gives…. • 1.74 X 10-5 = [H+]2 • 0.100 • [H+]2= 0.100 X 1.74 X 10-5 • [H+] = √(1.74 X 10-6) • [H+]= 1.319 X 10-3moldm-3 • = pH = 2.88 (using –log)

  14. Comparing strong and weak acids

  15. Strength vs. Concentration • Concentration is a measure of the amount of substance in a given volume of solution. • Strength is a measure of the extent to which an acid can donate H+. Measured as pKa values: • pKa = -log Ka

  16. Calculating the pH of a strong base.

  17. Calculating pH of a strong base. • To begin with we need to understand another term, Kw, this is the ionic product of water. • For water, Ka = [H+(aq)] [OH-(aq)] / [H2O(l)] • As water is always present in excess we ignore the term, [H2O(l)] to give: • Kw = [H+(aq)] [OH-(aq)] • At 298K Kw = 1x10-14 mol2 dm-6

  18. Calculating pH of a strong base. • Kw = [H+(aq)] [OH-(aq)] • At 298K Kw = 1x10-14 mol2 dm-6 • Kw = 1x10-14 = [H+]2 • Therefore [H+] = 1x10-7 • =pH7 at 298K, pH will fall as temperature increases.

  19. Calculating pH of a strong base. • What is the pH of a 0.1 mol solution of sodium hydroxide? • Kw = 1 x 10-14 = [H+] x [OH-] • Kw = 1 x 10-14 = [H+] x [0.1] • [H+] = 1 x 10-14 / 0.1 • [H+] = 1 x 10-13 • pH = -log [1 x 10-13] = 13

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