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Aqueous Equilibria Lecture 18 Kotz

# Aqueous Equilibria Lecture 18 Kotz - Topics Common Ion...

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1 Topics: Common Ion Effect From Salts and Partial Neutralization Buffer Systems: Resisting Changes in pH The Henderson-Hasselbalch Equation Acid-Base Titrations Titration Curves of Strong and Weak Acids & Bases Titration Curves of Polyprotic Acids Indicators: Endpoint vs. Equivalence Point Solubility Product of Slightly Soluble Salts (K sp ) Common Ion Effect Precipitation Reactions Revisited Solubility and Complex Ions Solubility, Ion Separations, and Qualitative Analysis The Common Ion Effect: The ionization of an acid or base is limited by the presence of its conjugate base or acid (based on Le Chatelier’s Principle) Can occur from: -The effect of a partial neutralization of an acid or base -The presence of a salt containing the conjugate of the acid or base Affects the equilibrium expression of the acid or the base. Example 1: What is the pH of the solution that results from adding 30.0mL of 0.100M NaOH to 45.0mL of 0.100M acetic acid? Moles NaOH: (.0300L)(0.100M) = .0030mol NaOH Moles HC 2 H 3 O 2 : (.0450L)(0.100M) = .0045 HC 2 H 3 O 2 .0030mol of the acid are neutralized leaving: .0030mol C 2 H 3 O 2 - (.04M) and .0015mol HC 2 H 3 O 2 (.02M) K a for acetic acid is 1.8x10 -5 1.8x10 -5 = (x)(.04+x) / (.02-x) x = 8.99x10 -6 = [H 3 O + ] pH=-log(8.99x10 -6 ) = 5.05 *Note that since x is small, the equation can be simplified to 1.8x10 -5 = (x)(.04/.02) and you get almost the same answer. Example 2: Assume you have a 0.30M solution of formic acid (HCO 2 H) and have added enough sodium formate (NaHCO 2 ) to make the solution 0.10M in the salt. Calculate the pH of the formic acid solution before and after adding sodium formate. Answer: pH before: 1.8x10 -4 = (x)(x) / (0.30 – x) x = 7.26x10 -3 pH = -log(7.26x10 -3 ) = 2.139 = 2.14 pH after: 1.8x10 -4 = (x)(x+0.10)/(0.30-x) x = 5.36x10 -4 pH = -log(5.36x10 -4 ) = 3.27 Buffer Systems Resists changes to pH when a strong acid or base is added Requirements: -An acid capable of scavenging added OH - ions and a base capable of scavenging added H 3 O + ions. -The acid and the base of the buffer must not react with each other Made from weak acids and salts of their conjugate bases or weak bases and salts of their conjugate acids Relies on the common ion effect 5mL of 0.10M HCl added to buffered and nonbuffered systems (containing bromphenol blue indicator). Note the lack of color change for the buffered system

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2 Buffers and the Henderson-Hasselbalch Equation 3 3 [ ][ ] [ ] [ ] [ ] [ ] [ ] log [ ] a a a H O A K HA K HA H O A A pH pK HA + + = = = + Henderson-Hasselbalch Equation How Buffer Systems Work For added strong acid: H + (aq)(from acid) + A - (aq)(from buffer) HA (aq)(undissociated) For added strong base: HA (aq)(from buffer) + OH - (aq)(from base) A - (aq) + H 2 O (l) The best range of pH buffering occurs when the desired pH (to be buffered) is very close to the pK a of the weak acid being used in the buffer system Buffers with sufficient molarity must be used so as not to “exhaust” the buffer. Example: Using an acetic acid/sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00? (K a for acetic acid is 1.8x10 -5 ) Answer: 5.00 = 4.74 + log(acetate ion / acetic acid) .26 = log (acetate ion / acetic acid) (acetate ion / acetic acid) = 10 .26 = 1.82 So you would need a 1.82:1 mol (molarity) ratio of
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