Ch09-TCF

Ch09-TCF - Quantitative Chemical Analysis Chapter 9:...

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Quantitative Chemical Analysis Quantitative Chemical Analysis Chapter 9: Monoprotic Acid-Base Equilibria Monoprotic Acid-Base Equilibria
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Strong Acids and Bases dissociate completely in water forming [H + ] or [OH ]. • Recall from Chapter 8 that 1.0×10 8 M HCl in water gives a pH of 6.98, not 8.00. • Recall autoprotolysis of water: K w = [H + ][OH ] = 1.01×10 14 at 25 °C. • Autoprotolysis must always form H + and OH in identical amounts . Thus, at pH = 3.00, [H + ] = 1.0×10 3 and [OH ] = K w /[H + ] = 1.0×10 11 so only 1.0×10 11 out of the 1.0×10 3 [H + ] can come from the autoprotolysis. So, the K w equilibrium can be ignored below pH ~ 6 and above pH ~ 8. Figure 9.1 pH as a function of concentration of strong acid or strong base in water.
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Weak Acids and Bases dissociate only partially when dissolved in water. Recall, K a · K b = K w and p K a + p K b = 14 @ 25 °C Now, let’s consider a specific example of a weak acid … benzoic acid at 0.100 M in water @ 25 °C C O OH C O OH + H 2 O + H 3 O + p K a = 4.202, K a = 6.28×10 5 We set up the concentration table …
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To repeat, What are the concentrations? What is the pH? K a = 6.28×10 5 = ( x ) 2 (0.100 x ) Assumption 1: ignore K w Assumption 2: x << 0.1 so, (0.100· K a ) 1/2 = x x = [H + ] = [OBz ] = 2.51×10 3 M pH = 2.600 3 Check assumption 1 : [OH ] = K w /[H + ] = 3.89×10 12 M, so only that amount of [H + ] comes from K w << 2×10 3 Check assumption 2 : Lets not ignore x wrt 0.100. 0 = x 2 + (6.28×10 5 ) x 6.28×10 6 and x = [H + ] = 2.47 5 ×10 3 (1.4% error) pH = 2.606 (0.2% error) Note also that 0.0025 is 2.5% of 0.100.
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α ) α Α [A ] [A ] + [HA] 0.00251 = 0.0251 (2.5% dissociation) 0.100 = [A ] F fromal concentration For 0.100 F benzoic acid, Fractional Composition
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This note was uploaded on 03/28/2011 for the course CHEM 300L at USC.

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Ch09-TCF - Quantitative Chemical Analysis Chapter 9:...

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