Fall 2011 lect 17 acid bases dissoc Chem 162 HOME

Fall 2011 lect 17 acid bases dissoc Chem 162 HOME - Exam II...

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1 Calculators of this type are not allowed for quizzes and the exams. This type is OK. Exam II Tuesday November 8, 2011 in class (lecture). 7:15 PM – 8:35 PM Pumpkins, Bats, Zombies, Skeletons
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Exam II Tuesday November 8, 2011 In Lecture Chapters 13, 14, 15 13.1 up to 15.11 See pages 527-664 Pumpkins, Bats, Zombies, Skeletons 2
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Pumpkins, Bats, Zombies, Skeletons 4 Lecture 17 Outline of Chapter 15 15.5 Polyprotic acids 15.6 Ions as acids and bases 15.7 Continue Buffers What is the pH of solutions of different salts? 15.9 Indicators 15.10 Titrations End Point vs Equivalence point 15.11 Lewis Acid-Bases Metal ions in solution give basic or acid solutions
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Pumpkins, Bats, Zombies, Skeletons 5 Key Points on Buffered Solutions 1.They are weak acids or bases containing a common ion. Examples: NaOAc + HOAc, acidic Example: NH 4 Cl + NH 3 , basic One of the ions should be from strong acid or strong base. 2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.
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Pumpkins, Bats, Zombies, Skeletons 6 Adding strong base Lets add a strong base 0.02M NaOH and see what happens to the pH. NaOH(aq) gives 0.02M OH - HOAc (aq) + OH - (aq) = H 2 O(l) + OAc - (aq) start 1M 1E-7 1M add 0.02(left out) change -0.02 -0.02 +.02 Equil 0.98 0 1.02
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Use H & H Equation for equilibrium concentrations pH = pKa + log [OAc-]/[HOAc]; pH = 4.74 + log(1.02/.98) = 4.76 was 4.77 when solved directly. Pumpkins, Bats, Zombies, Skeletons 7
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Pumpkins, Bats, Zombies, Skeletons 8 Buffer, adding acid Lets add acid: 0.02M HCl HCl(l) + H 2 O(l) = H 3 O + (aq) + Cl - (aq) 0.02M gives 0.02M each OAc - (aq) + H 3 O + (aq) = HOAc (aq) + H 2 O (l) start 1M 1M add 0.02 change -0.02 -0.02 +.02 Equil .98 0 1.02
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Use H & H Equation for equilibrium concentrations pH = pKa + log [OAc-]/[HOAc]; pH = 4.74 + log(0.98/1.02) pH = 4.72 same as exact solution Pumpkins, Bats, Zombies, Skeletons 9
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Pumpkins, Bats, Zombies, Skeletons 10 Easier way for base There has to be an easier way to calculate pH of basic buffers. B(aq) + H 2 O (l) = BH + (aq) + OH - (aq) K b =[BH + ][OH - ]/[B] There is an easier way to do these calculations. [OH - ] = K b [B]/[BH + ], take –logs of both sides: -log [OH - ] = -log K b -log[B]/[BH + ] reduces to because of definitions of pOH and pK b . pOH = pK b + log [BH + ]/[B]; Generalizing: pOH = pK b + log [CA, salt of the WB]/[WB] known as Henderson-Hasselbalch equation.
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Buffer Capacity and Buffer Range There is a limit to the ability of a buffer solution to neutralize added acid or base. This buffer capacity is reached before either buffer component has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal or nearly so. Therefore, a buffer is most effective when the desired pH of the buffer is very near p K a of the weak acid of the buffer.
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This note was uploaded on 11/09/2011 for the course GEN CHEM 162 taught by Professor Tavss during the Spring '11 term at Rutgers.

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Fall 2011 lect 17 acid bases dissoc Chem 162 HOME - Exam II...

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