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Solution Equilibria

# Solution Equilibria - Solution Equilibria I Strong acids...

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1 Solution Equilibria I. Strong acids and Bases—pH A. Self dissociation of water. 1. Recall that the following equilibrium takes place in aqueous solutions: 2H 2 O H 3 O + + OH K = K w = [H 3 O + ][OH ] = 1.0x10 –14 at room temperature. 2. This expression must be satisfied in all aqueous solutions, that is there is always H 3 O + and OH ions present in these solutions. Examples: a. In a 0.15 M HCl solution; [H 3 O + ] = 0.15 M, [OH ] = 1.0x10 –14 [H 3 O + ] = 1.0x10 –14 0.15 = 6.7x10 –14 M b. In a 0.054 M NaOH solution; [OH ] = 5.4x10 –2 M, [H 3 O + ] = 1.0x10 –14 [OH ] = 1.0x10 –14 5.4x10 –2 = 1.9x10 –13 M c. In a 0.15 M NaCl solution; [H 3 O + ] = [OH ] = 1.0x10 –7 M B. pH 1. pH = -log[H 3 O + ] where log = base 10 logarithm [H 3 O + ] = 10 -pH a. If [H 3 O + ] = 3.0x10 -4 , then pH = -log(3.0x10 -4 ) = -(-3.52) = 3.52 b. If pH = 5.75, then [H 3 O + ] = 10 -5.75 = 1.78x10 -6 M 2. In general, p before any variable means - log(variable). pOH = - log[OH - ] pMg = - log[Mg 2+ ] pK = - log K (K = the equilibrium constant) 3. Examples. a. Consider the following solutions of some strong acids and bases. Solution [H 3 O + ] [OH - ] pH pOH pH+pOH 0.10M HCl 1.0x10 -1 1.0x10 -13 1.0 13.0 14.0 0.01M HCl 1.0x10 -2 1.0x10 -12 2.0 12.0 14.0 0.005M HCl 5.0x10 -3 2.0x10 -12 2.3 11.7 14.0 0.1M NaCl 1.0x10 -7 1.0x10 -7 7.0 7.0 14.0 0.02M NaOH 5.0x10 -13 2.0x10 -2 12.3 1.7 14.0 b. Note

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2 a. [H 3 O + ][OH - ] = 1.0x10 -14 = K w b. pH + pOH = 14.0 = pK w c. An acidic solution = one in which pH < 7; [H 3 O + ] > 1.0x10 -7 , [OH - ] < 1.0x10 -7 . A neutral solution = one in which pH = 7; [H 3 O + ] = [OH - ] = 1.0x10 -7 . A basic solution = one in which pH > 7; [H 3 O + ] < 1.0x10 -7 , [OH - ] > 1.0x10 -7 . II. Weak acid/base equilibria. A. Equilibrium Constant Expressions. 1. Weak monoprotic acids. Consider acetic acid, HC 2 H 3 O 2 . It is a weak monoprotic acid. a. Equilibrium HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 - K a = [H 3 O + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] = 1.8x10 -5 at 25 ° C K a = acid dissociation constant b. Note. 1) [H 2 O] does not appear in the equilibrium constant expression. Since these are dilute aqueous solution reactions, the solvent H 2 O can be considered a "pure liquid" having a unit activity. 2) The equilibrium constant is very small. Therefore, at normal concentrations, only a very small percent of the acetic acid is dissociated, 2. Weak soluble bases. a. Consider ammonia, NH 3 . Due to the hydrolysis of NH 3 , an aqueous solution of NH 3 in water will be basic. The weak base ‘ammonium hydroxide” is just an aqueous solution of NH 3 . 1) Equilibrium NH 3 + H 2 O NH 4 + + OH - K b = [NH 4 + ][OH - ] [NH 3 ] = 1.8x10 -5 at 25 ° C 2) As was found for the acetic acid equilibrium, [H 2 O] does not appear in the equilibrium constant expression. 3) The reaction is called a hydrolysis reaction = reaction of a substance with water to give an acidic or basic solution where the H + or OH - comes from the H 2 O. b. The anions of the weak acids are the conjugate bases of the weak acids. Weak acids will yield fairly strong conjugate bases that can remove protons from water molecules, giving basic solutions. Therefore NaC 2 H 3 O 2 will dissolve in water to give a basic solution. The C 2 H 3 O 2 - will hydrolyze to give a basic solution.
3 1) Equilibrium C 2 H 3 O 2 - + H 2 O HC 2 H 3 O 2 + OH - K b = [HC 2 H 3 O 2 ][OH - ] [C 2 H 3 O 2 - ] = 5.6x10 -10 at 25 ° C 3. K a and K b for conjugate acid base pairs.

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