Chapter 16 Other Equilibria Week 1 2009

Chapter 16 Other Equilibria Week 1 2009 - CHAPTER 16 OTHER...

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CHAPTER 16. OTHER EQUILIBRIA Section 16.2: Common-Ion Effect ( example of Le Chatelier’s Principle) = the shift in an equilibrium caused by the addition (or removal) of one of the participating species in the equilibrium. Example 1 : add lactate ion (C 3 H 5 O 3 - ) to lactic acid (HC 3 H 5 O 3 ) solution HC 3 H 5 O 3 (aq) + H 2 O (l) C 3 H 5 O 3 - (aq) + H 3 O + (aq) if we add C 3 H 5 O 3 - SHIFT Net Effect : [H 3 O + ] decreases pH increases [OH - ] increases pOH decreases Example 2 : add NH 4 Cl to NH 3 solution. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) if we add NH 4 + SHIFT Net Effect : [OH - ] decreases, pOH increases [H 3 O + ] increases, pH decreases ------------------------------------------------------------------------------- Type of problem we must be able to solve for this Chapter is when we start with a weak acid and its conjugate base: Problem : Calculate [H 3 O + ] in a solution that is 0.10 M in HF and 0.20 M in NaF. Also calculate % ionization. ** First, we have to decide which type of equation we should use** : HF (aq) H + (aq) + F - (aq) (K a ) or F - (aq) + H 2 O (l) HF (aq) + OH - (aq) (K b ) 1
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Answer : Since both include HF, F - , H + , OH - (the last two are related by [H + ][OH - ] = 1.0 x 10 -14 ), we can use either equation! Let’s use: HF (aq) H + (aq) + F - (aq) [I] 0.10 M 0 0.20 M [C] -x +x +x [E] (0.10-x) x (0.20 + x) K a = [HF] ] ][F [H - + = 6.8 x 10 -4 = x) - (0.10 x) x(0.20 + 0.10 x 0.20 x = (0.10) (6.8 x 10 -4 )/0.20 = 3.4 x 10 -4 M (x << 0.10) [H + ] = [H 3 O + ] = 3.4 x 10 -4 M Compare : in pure 0.1 M HF, [H 3 O + ] = 8.5 x 10 -2 M in pure 0.2 M NaF, [H 3 O + ] = 6.0 x 10 -9 M % ionization of HF = [HF] ] [H + 100% = 0.10 ) 10 x (3.4 -4 100% = 0.34% Compare : For pure 0.1 M HF, % ionization is 8.5% ** The presence of NaF in the solution has decreased the % ionization of HF (equilibrium shifted to the left) less [H 3 O + ] less acidic. ** Calculations also involving volume changes: These are very common (especially in titrations), so get used to handling them ! See Example 16.3 in the textbook. Below is another example to help you some more. 2
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: What is the pH of a solution prepared from adding 60.0 mL of 0.100 M NH 3 to 40.0 mL of 0.100 M NH 4 Cl ? K b = 1.8 x 10 -5 for NH 3 . Note : The mixture involves the NH 3 /NH 4 + conjugate acid/base pair. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Conc’s : Since two solns are mixed, both have been diluted. We must calculate new concs. ** Use this: V i M i = V f M f ** V f = 60.0 mL + 40.0 mL = 100.0 mL [NH 3 ] f = V i M i /V f = (60.0 mL)(0.100 M)/100.0 mL = 0.0600 M [NH 4 + ] f = V i M i /V f = (40.0 mL)(0.100 M)/100.0 mL = 0.0400 M Now, we can calculate pH : NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) [I] 0.0600 - 0.0400 0 [C] -x +x +x [E] (0.0600-x) (0.0400+x) x K b = 1.8 x 10 -5 = x) - (0.0600 x) x(0.0400 + 0.0600 0.0400x x = (0.0400) )(0.0600) 10 x (1.8 -5 = 2.7 x 10 -5 M pOH = -log (2.7 x 10 -5 ) = 4.57 pH = 9.43
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This note was uploaded on 05/30/2010 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.

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Chapter 16 Other Equilibria Week 1 2009 - CHAPTER 16 OTHER...

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