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Ch10-TCF

Ch10-TCF - Quantitative Chemical Analysis Chapter 10...

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Quantitative Chemical Analysis Quantitative Chemical Analysis Chapter 10: Polyprotic Acid-Base Equilibria Polyprotic Acid-Base Equilibria
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Polyprotic Acids – Notation Consider a diprotic acid We have acidic equilibria … H 2 A H + + HA K a1 K 1 HA H + + A 2 K a2 K 2 And basic equilibria … A 2 + H 2 O OH + HA K b1 HA + H 2 O OH + H 2 A K b2 And, K a1 · K b2 = K w K a2 · K b1 = K w Note : NOT K a1 · K b1 = K w K a2 · K b2 = K w
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Example Case 1 : Let’s prepare a solution of H 2 Mal in pure water at 0.500 formal … C HC O OH C O Maleic Acid K a1 = 1.23 × 10 2 p K a1 = 1.910 K a2 = 4.66 × 10 7 p K a2 = 6.332 Since p K 2 p K 1 > 4, we assume K 2 is negligible We set up the concentration table … K 1 = x 2 = 1.23 × 10 2 (0.500– x ) 0 = x 2 + (1.23 × 10 2 ) x –6.15 × 10 3 x = 7.25 × 10 2 = [H + ] = [HMal ] and pH = 1.14 Ignoring x wrt 0.500 gives x = 7.84 × 10 2 (16% of 0.500) – too big to ignore x = 7.84 × 10 2 (approximate) has 8% error wrt 7.25 × 10 2 (quadratic)
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Was the assumption that HMal H + + Mal 2 is negligible warranted? Let’s take the above answer and check. K a2 = 4.66×10 7 = (0.073+ x )( x ) (0.073 x ) Assume x << 0.073, so x = 4.66×10 7 ( << 0.073 ) It is clearly warranted to ignore the second ionization when dealing with the first.
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Case 2 : Let’s prepare a solution of Na 2 Mal in pure water at 0.500 formal. Na 2 Mal is a strong electrolyte, so … Mal 2 + H 2 O HMal + OH K b1 K b1 = K w = 1.01 × 10 14 = 2.16 7 × 10 8 = x 2 K a2 4.66 × 10 7 (0.500 x ) (Ignoring x wrt 0.500) x = 1.47 × 10 4 = [OH ] = [HMal ] And [H + ] = K w = 1.01 × 10 14 = 6.87 × 10 11 [OH ] 1.47
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Ch10-TCF - Quantitative Chemical Analysis Chapter 10...

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