CH302 Chapter 8 answers

CH302 Chapter 8 answers - The Common Ion Effect Weak Acid...

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1 [ ][ ] [] () ( ) x x x + = × = = 1 . 0 2 . 0 10 8 . 1 COOH CH COO CH H K 5 3 - 3 + a The Common Ion Effect Weak Acid PLUS Salt of Weak Acid and Strong Base : We have a solution that is 0.10M CH 3 COOH and 0.2M NaCH 3 COO. We assume the salt breaks apart fully and the weak acid partially dissociates: assume v. little anion comes from rxn 2, most from rxn 1 We get x = 9.0 x 10 -6 . This is small enough to confirm the shortcut. Try this with JUST the 0.1M acetic acid and you find: The Common Ion Effect Weak Acid PLUS Salt of Weak Acid and Strong Base compared to just the Weak Acid: These much lower values are caused by the common ion effect. [ ] x x x + = × = = 2 . 0 1 . 0 10 8 . 1 NH OH NH K 5 3 - + 4 b The Common Ion Effect Weak Base PLUS Salt of Weak Base and Strong Acid : We have a solution that is 0.20M NH 3 and 0.1M NH 3 Cl. We assume the salt breaks apart fully and the weak base partially dissociates: assume v. little cation comes from rxn 2, most from rxn 1 We get x = 3.6 x 10 -5 . This is small enough to confirm the shortcut. 0.1M 0.1M 0.1M (0.2 – x) M x M x M The Common Ion Effect These much lower values are caused by the common ion effect. Weak Base PLUS Salt of Weak Base and Strong Acid : comparison to just the Weak Base : First, calculate the pH of the original buffer solution. NH 4 Cl NH 4 + + Cl - All the salt falls apart to give 0.200M of both ions. Cl - wont do anything so we worry about NH4+; and include it in our ICE table for the ammonia: NH 3 + H 2 O NH 4 + + OH - 0.1M / 0.2M ~0M a smal amount of ammonia wil react: -x + x + x (0.1-x)M (0.2+x)M xM Ka = 1.8 x 10 -5 = [NH 4 + ]+ [OH - ] = (0.2+x)M xM [ NH 3 ] (0.1-x)M 1.8 x 10 -5 = (0.2) x gives x = 9 x 10 -6 M since x = [OH - ] (0.1) pH = 1x10 -14 / 9 x 10 -6 = 1x10 -14 = 1.11 x 10 -9 pH = 8.95 Buffering Action: Weak Base Buffer Next, calculate the concentration of all species after the addition of the 0.02 moles of gaseous HCl. ± The HCl will react with some of the ammonia and change the concentrations of the species. ± This is another limiting reactant problem. mol 0.220 mol 0.080 mol 0 rxn. After mol 0.020 + mol 0.020 - mol 0.020 - Change mol 0.200 mol 0.100 mol 0.020 Initial Cl NH NH HCl 4 3 + Note these are all in moles, we had 1L of solution
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2 Buffering Action : Weak Base Buffer HCl NH NH Cl Initial 0.020 mol 0.100 mol 0.200 mol Change - 0.020 mol - 0.020 mol + 0.020 mol After rxn. 0 mol 0.080 mol 0.220 mol mol 1.0 L 34 NH 3 4 +→ == MM 0080 0220 0 220 . . . . We now have the new concentrations of ammonia and ammonium ion Action of adding the acid: Using the new conc of the ammonia and ammonium ion and redo the equilbrium calc: NH 3 + H 2 O NH 4 + + OH - 0.08M / 0.22M ~0M a smal amount of ammonia wil react: -x + x + x (0.08-x)M (0.22+x)M xM Ka = 1.8 x 10 -5 = [NH 4 + ]+ [OH - ] = (0.22+x)M xM [ NH 3 ] (0.08-x)M 1.8 x 10 -5 = (0.22) x gives x = 6.5454 x 10 -6 M since x = [OH - ] (0.08) pH = 1x10 -14 / 9 x 10 -6 = 1x10 -14 = 1.5277 x 10 -9 pH = 8.81 ie only dropped 0.14 units. Buffering Action : Weak Base Buffer Now see what would happen if we added some base: If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.
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CH302 Chapter 8 answers - The Common Ion Effect Weak Acid...

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