# lec19 - –(aq • Can calculate K b from K a using HOAc(aq...

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1 Last Time • Dilute Solutions of Acids and Bases [Acid,Base] < 10 –6 M •Po l yp ro t i cac ids – Diprotic – H 2 S – Triprotic – H 3 PO 4 Example Calculate the concentration of all components of a 0.01 M H 3 PO 4 solution given that K a1 = 7.6 x 10 –3 , K a2 = 6.2 x 10 –8 , and K a3 = 2.1 x 10 –13 . So far we have … • Dissolved weak acids and bases in H 2 O • Acids: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) K a = • Bases: B (aq) + H 2 O(l) HB(aq) + OH (aq) K b = [H 3 O + ][A ] [HA] [HB][OH ] [B ] Ions as Acids • All conjugate acids of weak bases NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) • Can calculate K a from K b using NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) K a = = = [NH 3 ][H 3 O + ] [NH 4 + ] [NH 4 + ][OH - ] [NH

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2 Ions as Bases • All conjugate bases of weak acids OAc (aq) + H 2 O(l) HOAc(aq) + OH
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Unformatted text preview: – (aq) • Can calculate K b from K a using HOAc(aq) + H 2 O(l) → H 3 O + (aq) + OAc – (aq) K b = = = [HOAc][OH – ] [OAc – ] [OAc – ][H 3 O + ] [HOAc] K w K w K a ← ← Example What is the pH of a 3.2 x 10 –2 Mg(OAc) 2 solution given that K a = 1.8 x 10 –5 for HOAc? Combining Acids, Bases, & Salts • Considered simple weak acids and bases • Extend equilibrium concepts to mixtures – Weak acids and salts – use K a – Weak bases and salts – use K b – Salt modifies initial concentrations Example Calculate the pH of a solution that is 0.25 M in CH 3 NH 2 and 0.32 M in CH 3 NH 3 Cl using Kb = 3.6 x 10 –4 ....
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lec19 - –(aq • Can calculate K b from K a using HOAc(aq...

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