Workshop10 - N am es_3 0 ptS total CHEM 104 Workshop 10...

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Unformatted text preview: N am es:_,__3 0 ptS total ___________________ ____ ________________ __ CHEM 104 Workshop 10: Buffers A buffer consists ofa solution containing both a weak acid and it’s conjugate base. Lets try and solve some old exam problems: (30 points] 1. Consider a 500 ml buffer that contains 0.12M HOCl and 0.08M KOCl (K3 for HOCl= 3.5 X 108] What is the initial pH of this buffer? HINT: Remember the Henderson-hasselbach eqn [5 pts] pH = pKa + log [Base]/[Acid] = -log[3.5x10-8] + log (0.08/0.12) = 7.45 + [-0.176] = 7.27 2. Now, let’s add 2 ml of 10 M KOH to this buffer and ask what species are present in the buffer after the addition of the strong base? (12 pts) HINT: This is a limiting reagent problem in which the strong base reacts with the weak acid component of the buffer. The reaction will go to completion, so all the limiting reagent will be consumed. The net chemical reaction you are interested in is: HOCl + 0H" —) 0C1" + H20 To approach this problemI follow the following 3 step procedure: ' Step 1: determine the number of moles of HOCl, OH" and 0C1" that you begin with. (Rememb er that you must work in moles, not molarity, for a limiting reagent problem) ' Step 2: determine the limiting reagent and calculate the number of moles of HOCl, OH" and OCl' that are present in the solution after the reaction. [Remember that one of these will be completely consumed as the limiting reagent) ' Step 3: calculate the concentrations of these species in units ofmolarity after reaction. Step 1: Moles before reaction HOCl -) 0.12 M HOCl * 0.5 L = 0.06 moles OH" -) 10M KOH * 0.002 L = 0.02 moles 0C1" -) 0.08 M KOCl * 0.5 L = 0.04 moles Step 2: LR and moles after reaction OH- is the LR and will be completely consumed when it reactions with HOCl. The concentration of HOCl will decrease and the concentration of OCl' will increase as follows: HOCl OH" DC]— I 0.06 0.02 0.04- C -0.02 -0.02 +0.02 E 0.04- 0 0.06 Step 3: Concentrations after reaction [HOCl] = 0.040 moles/ 0.502 L= 0.080 M [OCl'] = 0.060 moles /0.502 L = 0.120 M Nalnes:____30 ptS total__ 3. What is the pH of the buffer after the addition of 2ml of 10M KOH? By how many units did the pH of the buffer solution change? HINT: Use the information you found in problem 2. (5 pts) pH = pKa + log [Base]/[Acid] = -log[3.5x10-8) + log (0.12/0.08) = 7.45 + [0.176) = 7.63 ApH = 7.63 — 7.28 = 0.35 units 4. Now let’s compare this to the addition of the same amount of OH" to an unbuffered solution. a) What is the pH of500ml of pure water? (1 pt) No calculation needed here -) 7 b) What is the pH ofthe solution after 2 ml of 10 M KOH has been added? [5 pts) 10 M KOH * 0.002 L = 0.02 moles KOH 0.02 moles/0.502 L = 0.04 M OH" in the new solution pOH = -log [0.04] 21.39 pH = 14 — 1.39 = 12.6 ~13 c] By how many units did the pH of the unbuffered solution change? (2 pt] ApH = 13 — 7 = 6units EXTRA CREDIT: | 5 ptsl Hopefully you found that the pH of the buffered solution changed less than the pH ofthe unbuffered solution. Why didn't the pH of the buffered solution change as much? What happened to the OH" you added to the buffered solution? Big hint: Look at the K3 equilibrium expression for the following reaction: HOCl[aq] + H20[l) H30+(aq) + 0Cl'(aq) and think in terms of LeChatelier’s principle -) what would happen if you added OH" to this reaction. If you add hydroxide to this reaction mixture, it will react with the H30, resulting in a shift of the reaction to the right. In other words, the OH" you add will almost completely be consumed in converting HOCl to 0C1". As a result, the pH will not change very much. ...
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