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Recap – Last Lecture Intermolecular forces occur between molecules. 1 ForceAttractionEnergy/kJ mol -1 Example Dispersion forces Fluctuations in e - cloud0.05.

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Presentation on theme: "Recap – Last Lecture Intermolecular forces occur between molecules. 1 ForceAttractionEnergy/kJ mol -1 Example Dispersion forces Fluctuations in e - cloud0.05."— Presentation transcript:

1 Recap – Last Lecture Intermolecular forces occur between molecules. 1 ForceAttractionEnergy/kJ mol -1 Example Dispersion forces Fluctuations in e - cloud0.05 - 40All molecules Dipole-dipole forces Partial charges5 - 25H-Cl Hydrogen- bonding X :::::::: H-X10 - 40H2OH2O

2 Acids and Bases Arrhenius definition of acids and bases (1884): An acid releases H + ions when dissolved in water A base releases HO – ions when dissolved in water Br  nsted – Lowry definition of acids and bases (1923): An acid is a proton donor A base is a proton acceptor 2 Note: H + (aq) and H 3 O + (aq) are used interchangeably.

3 Conjugate pair A conjugate acid – base pair differ by H +. For example NH 4 + is the conjugate acid of NH 3 and NH 3 is the conjugate base of NH 4 +. 3

4 pH scale [H + ] is used as a measure of acidity and for convenience a log scale is often employed: pH = – log [H + (aq)] In neutral solution [H + ] = [OH – ] = 1 x10 –7 and pH = 7 at 25  C In acid solution [H + ] >1 x10 –7 and [OH – ] <1 x10 –7 and pH < 7 In basic solution [H + ] 1 x10 –7 and pH > 7 4

5 Example Calculate the pH of a 0.002 M solution of sulfuric acid. H 2 SO 4 → 2H + + SO 4 2- (one mole of H 2 SO 4 dissociates to give two moles of protons)  [H + ] = 0.004 M  pH = – log (0.004) = 2.4 (NB number of significant figures is given by the mantissa of log term.) 5

6 Autoionisation of Water H 2 O(l) ⇌ H + (aq) + OH – (aq) K w = [H + (aq)][OH – (aq)] = 1 x 10 -14 at 25  C pK w = -log K w = -log (1 x 10 -14 ) = 14.0 pH and pOH may be interconverted using pK w consequently it is convention to report all solutions – both acidic and basic ones in terms of pH. pK w = pH + pOH = 14 6

7 Example Calculate the pH of a 0.002 M solution of sodium hydroxide. NaOH → Na + + OH - (one mole of NaOH dissociates to give one mole of hydroxide ions)  [OH - ] = 0.002 M  pOH = – log (0.002) = 2.7  pH = 14 – 2.7 = 11.3 7

8 Strong & Weak Acids & Bases The terms strong and weak have a specific meaning in an acid – base context. Strong indicates complete dissociation into ions, e.g. HNO 3 → H + (aq) + NO 3 – (aq) Ca(OH) 2 → Ca 2+ (aq) + 2 OH – (aq) and the concentration of the ions, and hence pH, is obtained directly from the amount of starting material. Common strong acids are: HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 Common strong bases are:NaOH, KOH, Ca(OH) 2, Ba(OH) 2 8

9 Strong & Weak Acids & Bases Weak indicates an equilibrium exists between the ions and the undissociated compound in solution with, very often, the undissociated compound dominating. The pH of this solution may only be calculated if the equilibrium constant, K, as well as the concentration of the starting material is known. e.g. CH 3 COOH(aq) ⇌ CH 3 COO – (aq) + H + (aq) NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH – (aq) Common weak acids include: HF, HNO 2, H 3 PO 4, CH 3 COOH and other carboxylic acids Common weak bases include: F –, NO 2 –, CH 3 COO –, NH 3 and most organic amines 9

10 Strong & Weak Acids & Bases There is a direct relationship between the strength of an acid and its conjugate base. So a strong acid (e.g. HCl) is completely dissociated in water and its conjugate base (Cl – ) is a very, very weak conjugate base and shows no tendency to accept a proton. 10 HCl → H + + Cl – and Cl – + H 2 O ––X → HCl + OH – (no reaction)

11 Strong & Weak Acids & Bases A weak acid (e.g. CH 3 COOH) is in equilibrium with its ions in water and its conjugate base (CH 3 COO –, a weak base) is also in equilibrium in water. CH 3 COOH ⇌ CH 3 COO – + H + and CH 3 COO – + H 2 O ⇌ CH 3 COOH + OH – 11

12 Quantification The position of equilibrium of a weak acid is given by the value of K a. e.g. CH 3 COOH(aq) ⇌ CH 3 COO – (aq) + H + (aq) K a = [CH 3 COO – ][H + ] / [CH 3 COOH] For convenience the values are usually recorded as pK a values where pK a = -log K a. The stronger the acid the smaller the pK a value (and the larger K a value) 12

13 Quantification The position of equilibrium of a weak base is given by the value of K b. e.g. NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH – (aq) K b = [NH 4 + ][OH – ] / [NH 3 ] Once again, the values are usually recorded as pK b values. The stronger the base the smaller the pK b value (and the larger K b value) (Recap: pH + pOH = pK w = 14 but also pK a + pK b = pK w = 14) 13

14 pH and pK a In the same way it is convenient to report both acid and base concentration on the same scale (pH), data books often report weak base strength in terms of the pK a (of the conjugate acid). 14

15 15 Applications Many pharmaceuticals are composed of molecules with acidic or basic groups. When in their charged form, this may aid administration as they are likely to be water soluble. In particular drugs containing an amine (nitrogen) group are often sold as the hydrochloride salt. This not only increases water solubility by also hinders the oxidation of the amine group and hence extends shelf life of the product. Sold as a hydrosulfate and used for asthma relief (eg ventolin) Sold as a chloride and used for hayfever and allergy relief (eg zyrtec)

16 By the end of this lecture, you should be able to: −list common acids and bases. −define acids and bases according to the Arrhenius and Bronsted-Lowry models. −use the definitions of pH and K w to quantify the acidity and basicity of aqueous solutions. −explain the difference between a strong and weak acid and the difference between a strong and weak base in terms of the percentage dissociation in solution. −be able to complete the worksheet (if you haven’t already done so…) 16 Learning Outcomes:

17 17 Questions to complete for next lecture: 1.Consider 1 M water solutions of NaOH, H 2 SO 4, H 3 PO 4 (pK a1 = 2.15), HF (pK a = 3.17) and CH 3 COOH (pK a = 4.76). Arrange these solutions in order of increasing pH. 2.Calculate the pH of the following solutions: a.0.015 M HCl b.0.0015 M HCl c.0.03 M KOH d.The solution formed from mixing 250 mL of 0.03 M HNO 3 with 150 mL of 0.02 M Ca(OH) 2.


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