hw5 soln - CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS...

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19-1 CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS FOLLOW-UP PROBLEMS 19.1 Plan: The problems are both equilibria with the initial concentration of reactant and product given. For part (a), set up a reaction table for the dissociation of HF. Set up an equilibrium expression and solve for [H 3 O + ], assuming that the change in [HF] and [F ] is negligible. Check this assumption after finding [H 3 O + ]. Convert [H 3 O + ] to pH. For part (b), first find the concentration of OH added. Then, use the neutralization reaction to find the change in initial [HF] and [F ]. Repeat the solution in part (a) to find pH. a) Solution: Concentration ( M ) HF( aq ) + H 2 O( l ) ' F ( aq ) + H 3 O + ( aq ) Initial 0.50 0.45 0 Change - x + x + x Equilibrium 0.50 - x 0.45 + x x Assumptions: 1) initial [H 3 O + ], from water, at 1.0 x 10 -7 M , can be assumed to be zero and 2) x is negligible with respect to 0.50 M and 0.45 M . [ H 3 O + ] = [ ] a HF K F   = () [ ] [] 4 0.50 6.8 x 10 0.45 = 7.5556 x 10 -4 = 7.6 x 10 -4 Check assumptions: 1) 1.0 x 10 -7 << 7.6 x 10 -4 Thus, the assumption to set the initial concentration of H 3 O + to zero is valid. 2) Percent error in assuming x is negligible: (7.5556 x 10 -4 /0.45) x 100 = 0.17%. The error is less than 5%, so the assumption is valid. Solve for pH: pH = -log (7.5556 x 10 -4 ) = 3.12173 = 3.12 Check: Since [HF] and [F ] are similar, the pH should be close to p K a , which equals log (6.8 x 10 -4 ) = 3.17. The pH should be slightly less (more acidic) than p K a because [HF] > [F ]. The calculated pH of 3.12 is slightly less than p K a of 3.17. b) Solution: What is the initial molarity of the OH - ion? NaOH mol 1 OH mol 1 NaOH g 00 . 40 NaOH mol 1 L NaOH g 40 . 0 = 0.010 M OH - Set up reaction table for neutralization of 0.010 M OH (note the quantity of water is irrelevant). Concentration ( M ) HF( aq ) + OH ( aq ) F ( aq ) + H 2 O( l ) Before addition 0.50 0.45 Addition 0.010 Change - 0.010 - 0.010 + 0.010 After addition 0.49 0 0.46 Following the same solution path with the same assumptions as part (a): [ H 3 O + ] = [ ] a HF K F = [ ] 4 0.49 6.8 x 10 0.46 = 7.2434782 x 10 -4 = 7.2 x 10 -4 M Check assumptions: 1) 7.2 x 10 -4 >> 1.0 x 10 -7 , the assumption is valid. 2) (7.2 x 10 -4 /0.46)100 = 0.16%, which is less than the 5% maximum assumption acceptable. Solve for pH: pH = -log (7.2434782 x 10 -4 ) = 3.14005 = 3.14 Check: With addition of base, the pH should increase and it does, from 3.12 to 3.14. However, the pH should still be slightly less than p K a : 3.14 is still less than 3.17.
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19-2 19.2 Plan: Sodium benzoate is a salt so it dissolves in water to form Na + ions and C 6 H 5 COO ions. Only the benzoate ion is involved in the buffer system represented by the equilibrium: C 6 H 5 COOH( aq ) + H 2 O( l ) ' C 6 H 5 COO ( aq ) + H 3 O + ( aq ) Given in the problem are the volume and pH of the buffer and the concentration of the base, benzoate ion. The question asks for the mass of benzoic acid to add to the sodium benzoate solution. First, find the concentration of C 6 H 5 COOH needed to make a buffer with a pH of 4.25. Multiply the volume by the concentration to find moles of C 6 H 5 COOH and use the molar mass to find grams of benzoic acid.
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hw5 soln - CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS...

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