1. Monday, November 16 Acids and Bases Chem 03 Fall 2009

2. Brønsted-Lowry Acids An acid is a proton (H + ) donor Strong acids totally dissociate in water –hydrochloric acid HCl12 M –nitric acid HNO 3 16 M –sulfuric acid H 2 SO 4 18 M Weak acids partially dissociate in water –HA ⇆ H + + A – –pK a measures the strength of the weak acid, pH at which HA = A –, or half of the H + comes off. –acetic acidHC 2 H 3 O 2 pKa = 4.75 –carbonic acidH 2 CO 3 pKa= 6.37 strong weak

3. Brønsted-Lowry Acids (proton donors) Strong Acids: completely dissociate HCl → H + + Cl - Weak Acids: only partially dissociate HA ⇆ H + + A - Ka = [products]/[reactants] = [H + ][A - ]/[HA] Identifying (page 654 15-2 and 657 15-3 of Chang) What do structures of the weak acids above have in common? pH scale: -log [H + ] elementary texts will say limit is 0-14, but this is not true: for example: strong acid HCl at 10 M, pH = -1

4. Br ø nsted-Lowry Bases A base is a proton (H + ) acceptor Strong bases totally dissociate in water –sodium hydroxide NaOH –barium hydroxideBa(OH) 2 –potassium hydroxide KOH Weak bases do not totally dissociate –sodium carbonateNa 2 CO 3 –ammonium hydroxideNH 4 OH –waterHOH weak strong

5. Neutralization Reactions Acid + Base  Salt + Water + heat HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) + heat –Base in burette is added to acid in flask until the moles of base added is equal to the moles of acid, this is called the equivalence point –moles acid = moles base: M a V a = M b V b –In practice, an indicator is used to monitor pH, when indicator changes color, the end point of titration is reached, the volume of base added is measured and the assumption is that you are very close to equivalence point Titration Reaction: often used to calculate M of either acid or base Figures from: http://www.csudh.edu/oliver/demos/carbnate/carbnate.htm

6. Calculate the volume of 0.1500 M NaOH base needed to neutralize 25.00 mL of 0.3000 M HCl When the base neutralizes the acid, there is an equivalent number of moles of acid and base –Moles acid = Moles base –M a V a = M b V b –0.3000 moles/L x 0.3000 L = 0.1500 moles/L x V b –Vb = Example of Titration Reaction I

7. Calculate the volume of 0.1500 M NaOH base needed to neutralize 25.00 mL of 0.3000 M HCl When the base neutralizes the acid, there is an equivalent number of moles of acid and base –Moles acid = Moles base –M a V a = M b V b –0.3000 moles/L x 0.3000 L = 0.1500 moles/L x V b –Vb = Example of Titration Reaction I –0.05000 L base or 50.00 mL

8. Calculate the pH at the start (25.00 mL of 0.3000 M HCl) and the end (after 50.00 mL of base added) Calculation of pH –START pH = -log [H+]= -log [0.3000] = –END pH = 7 at equivalence point for a strong acid-strong base titration. If you used phenolphthalein as indicator, would you have reached the endpoint? –(endpoint: a color change in indicator) –NO, PPN changes from clear to pink at pH 8.0 Example of Titration Reaction II

9. Calculate the pH at the start (25.00 mL of 0.3000 M HCl) and the end (after 50.00 mL of base added) Calculation of pH –START pH = -log [H+]= -log [0.3000] = –END pH = 7 at equivalence point for a strong acid-strong base titration. If you used phenolphthalein as indicator, would you have reached the endpoint? –(endpoint: a color change in indicator) –NO, PPN changes from clear to pink at pH 8.0 Example of Titration Reaction II 0.5229

10. Autoionization of Water H 2 O ⇆ H + + OH - K = [H + ][OH - ]/[H 2 O] but concentration of [H 2 O] is so HIGH Autoionization is just a drop in the bucket, so [H 2 O] stays ≈ constant at 55.5 M K x 55.5 = K w K w = [H + ][OH - ] = 1 x 10 -14 Take [– log] of right and left hand sides of equation: -log [H+][OH-] = - log [1 x 10 -14 ] Right hand side = (– log 1) + (– log 10 -14 ) = 14 Left hand side (– log H + ) + (– log OH - ) = pH + pOH pH + pOH = 14 This is always true, you can always get the concentration of H+ or OH- by knowing just one, and using the equation above. pH = 1pOH ____OH - ____ pH = 5pOH ____OH - ____ pH = 7pOH ____OH - ____ So, at pH = 7, H + = OH - = 10 -7 solution is neutral

11. Example of Titration Reaction III In previous example, calculate the pH of the solution if you added 0.50 mL too much base At equivalence, you have 75.00 mL of salt solution (25.00mL + 50.00 mL) at pH 7.0 The problem now becomes a dilution where you must first find out the molarity of base in the beaker, and then pOH, and then pH –First find the Molarity of Base in Beaker M 1 V 1 = M 2 V 2 (0.1500 M x 0.50 mL) = (M 2 x 75.50 mL) = 1.000 x 10 -3 M NaOH –Then find the pOH pOH = -log[OH-] = 3.000 –Finally find the pH pH = 14 – pOH = 11

12. Acid/Conjugate Base Pairs Table I Common NameChemical NameTypeAcid/ Conj. BaseKapKa Stomach AcidHClSHCl/Cl - ∞ Aspirinacetylsalicylic acidWC 9 H 8 O 4 / C 9 H 7 O 4 -1 3.0 x 10 -4 Lemon juicecitric acidWC 6 H 8 O 7 / C 6 H 7 O 7 -1 7.1 x 10 -4 Vinegaracetic acidWC 2 H 4 O 2 / C 2 H 3 O 2 -1 1.8 x 10 -5 Vitamin Cascorbic acidWC 6 H 8 O 6 / C 6 H 7 O 6 -1 8.0 x 10 -5 Baking Sodasodium bicarbonateWHCO 3 - /CO 3 -2 5.0 x 10 -7

13. Buffers resist changes of pH pKa = -log Ka So, if you know the Ka, the acid dissociation constant, you can calculate a pKa. Go back to Table I and fill in the pKa of the various acids, what do you notice about how the strength varies with the pKa? Now hold on for something even better: HA ⇆ H + + A - Ka = [H + ][A - ]/[HA] -log Ka = -log {[H+][A-]/[HA]} pKa = -log[H+] + {-log[A-]/[HA]} pKa = pH – log ([A-]/[HA])

14. Buffers pKa = pH – log ([A-]/[HA]) What happens when [A-]/[HA] = 1 __________________________________________________________ If you have a mixture of a weak acid HA and its conjugate base A-, approx equal concentrations, and at a pH close to the pKa, then you have a situation where even if you add a strong acid directly to the solution, the pH won’t change much because the HA/A- ratio will adjust itself and stay close to 1. This is called a BUFFERED solution A buffered solution is one which _________________________________________________