Consider a 1.0 L solution of 0.10 M acetic acid. Acetic acid is a weak acid only a small percent of the weak acid is ionized

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1 Chemistry 12 Acid- Base Equilibrium V Name: Date: Block: 1. Buffers 2. Hydrolysis Buffers An acid- base buffer is a solution that resists changes in ph following the addition of relatively small amounts of a strong acid or base. Example: Consider a 1.0 L solution of 0.10 M acetic acid. CH 3COOH (aq) + H 2O (l) CH 3COO - (aq) + H 3O + (aq) Acetic acid is a weak acid only a small percent of the weak acid is ionized Visually If a strong base was added to a solution, acetic acid will be there to neutralize the base. If a strong acid was added, there would be no species to neutralize it. In order for a buffer solution to be effective, equivalent concentrations of a weak acid and a conjugate base must be in solution. Circle the pairs of chemical species below that could be used to prepare a buffer solution: HNO 3 and NaNO 3 KF and HF HNO 2 and HNO 3 HCOOH and LiHCOO NaHSO 4 and Na 2SO 4 K 2CO 3 and K 2C 2O 4 HCl and NaCl KH 2PO 4 and K 2HPO 4

2 Acidic Buffer: Because Then Or in general The hydronium ion concentration (and therefore the ph) or a buffer solution depends on: 1. the K a value 2. the ratio of the concentration of the weak acid to its conjugate base Using our acetic acid example We now add 0.10 mol HCl to 1.0 L buffer solution with no volume change. The 0.10 mol H 3O + will be consumed [CH 3COO - ] will [CH 3COOH] will

3 We now add 0.10 mol OH - to 1.0 L buffer solution with no volume change. The 0.10 mol OH - will be consumed [CH 3COO - ] will [CH 3COOH] will Basic Buffer: Consider a NH 3/NH 4 + buffer solution: Because Then Or in general (Remember: K b for NH 3 = K w / K a for NH 4 +) Using our NH 3/NH 4 + example [OH - ] =

4 We now add mol HCl to 1.0 L buffer solution with no volume change. The mol H 3O + will be consumed [NH 3] will [NH 4 +] will [OH - ] = We now add mol OH - to 1.0 L buffer solution with no volume change. The mol OH - will be consumed [NH 3] will [NH 4 +] will [OH - ] = Hebden Workbook Pg. 181 #

5 Hydrolysis In previous Chemistry courses, you have learned about neutralization reactions where: Acid + Base à Salt + Water The salt produced in neutralization reactions are actually acidic or basic. The ions that make up the salt behave as weak acids or bases. Hydrolysis is the reaction of an ion with water to produce either: The conjugate base of the ion and hydronium ions or The conjugate acid of the ion and hydroxide ions. Circle the ions in the following list that represent cations of strong bases: Al 3+ Rb + Fe 3+ Cr 3+ Ca 2+ Sn 4+ Cs + Ba 2+ Circle the ions in the following list that represent the conjugate bases of strong acids: F - ClO 2 - ClO 4 - SO 4 2- Cl - NO 2 - CH 3COO - CN - NO 3 - Circle the following salts whose ions will not hydrolyze when dissociated in water. NH 4Cl Na 2CO 3 RbClO 4 Li 2SO 3 BaI 2 NH 4HCOO KIO 3 CsF CaBr 2 For the above list, for the salts that will hydrolyze: a) Write out the balanced dissociation equation. b) For the ion that will hydrolyze, write out the equation that will occur when it reacts with water. c) Write out the K a or K b expression. (the larger K a will win!)

6 Example: A 9.54g sample of Mg(CN) 2 is dissolved in enough water to make ml of solution. Calculate the ph of this solution. What is the concentration of Mg(CN) 2? What is initial concentration of each ion? (*Hint dissociation equation required) What are the 2 ions produced? Which will hydrolyze? What is the equation when it reacts with water? What is the base ionization expression (K b)? Make an ICE table. What is the [OH - ] at equilibrium? Calculate ph and poh.

7 Sodium carbonate is used in the manufacture of glass and also as a laundry additive to soften water. A ml aqueous solution of 0.50 M Na 2CO 3 is diluted to ml. Calculate the ph of the resulting solution.

8 The K b for pyridine, C 5H 5N, a weak base, is 4.7 x Calculate the ph of a 0.10 M solution of C 5H 5NHNO 3. Hebden Workbook Pg 148 #69-73