1 / 19

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.

olive
Télécharger la présentation

Starter question.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related