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Lecture

23polyproticacids.pdf

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Department
Chemistry
Course
ENCH 213
Professor
Diane Beauchemin
Semester
Fall

Description
Polyprotic acid-base equilibria • Diprotic acids and bases • Polyprotic acids and bases • Diprotic buffers • Fractional composition • Isoelectric and isoionic pH pH of diprotic acids and bases Example: amino acid → K >>a1 a2 + Solution of H A 2 • H 2 + H O ⇔2HA+ H O 3 + ca-x x x at equilibrium • Calculate as for weak monoprotic acid with + + 2 • Ka = Ka = 1H O ][3A]/[H A ] = x 2(c -x) a + • [HA] = [H O 3 = x = + • and [H A2] = c – a - • whereas [A] = = K a2 K a1 • even if K =a2 /10 a1 ◦ only 4% error in [H O ] calculated bu ignoring second ionization 0> error in pH of 0.01 3 HA+ H O ⇋ H O +A + - K - 2 3 - a2 A + H O 2 HA+ OH K b1 2- What is the pH and M concentration of a solution containing 0.100 M H M (for which pK =21.91 and 1 pK 2 6.33)? - + 2- -2 H 2 + H O ⇋2HM + H O 3 a) pH = 1.53 and [M ] = 2.95 x 10 M 0.100-x x x b) pH = 1 53 and [M ] = 4.68 x 10 M -7 - + 2 -1.91 2- -4 K 1 [HM][H O ]/[3 M] = x 20.100 – x= 10 = 0.0123 c) pH = 3.89 and [M ]= 1.29 x 10 M Solvingquadratic:x= 0.02946 M pH = 1.54 M = 10 -6.33= 4.68x10 cando this because 5 orders ofmag betweenK and K 1 2 Solution of A - - • A + H 2 ⇔ HA + OH- • cb-x x x at equilibrium • calculate as monobasic species with - - 2 • K b K =b1 /K w [Oa2[HA]/[A]=x /(c -x) b • [HA] = [OH] and [A] = c -x b • whereas [H A 2 = K [OH]b2HA]=K <> K and K << c a2 HA w a1 HA b) K a2 HA << K awd K << a1 HA c) K c >> K and K << c a1 HA w a2 HA d) K a1 HA >> K awd K >> a2 HA The amino acid glycine (H NCH COO2) has 2K for -COOH = 2.350aand pK for -NH = 9.778. a 3 At pH 6.06, which compounds or ions predominate in solution? - a) Amixture of H NCH C3OH and2H NCH COO 3 2 b) Amixture of H NCH C3O and 2 NCH COO 2 2 2- 6.06 is exactlythe sum - c) H N3H COO 2nly (onlyamphiprotic form) Polyprotic system • Example: triprotic system • H 3+ H O ⇔ 2 O + H A3 + 2 - K =a1 1 - + 2- • H 2 + H O ⇔ 2 O + HA 3 K =a2 2 2- + 3- • HA + H O ⇔ H2O +A 3 K =a3 3 • A + H O ⇔2HA + OH 2- - K =b1 /K w a3 2- - - • HA + H O ⇔ H2A + OH 2 K =b2 /K w a2 - - • H 2 + H O ⇔ 2 A+ OH 3 K =b3 /K w a1 Polyprotic system • H A: treat as weak monoprotic acid with K = K 3 a a1 • H 2 → intermediate i.e. amphiprotic - 2- ◦ H A⇔ 3 A ⇔ HA 2 ◦ • HA also intermediate (of H A andA )2 - 3- ◦ 3- • A : treat as monobasic with K = K = Kb/K b1 w a3 Diprotic buffers • treated as monoprotic using either: ◦ pH = pK + log1 [HA]/[H A] ) 2 - - ◦ or pH = pK +
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