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Chapter 18-again

Chapter 18-again - NaCH 3 COO – NaOH CH 3 COOH(will...

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Example 18-14 : Polyprotic Acids -many weak acids contain two or more acidic hydrogens -H 2 PO 4 and H 3 AsO 4 -calculation of equilibria for polyprotic acids is done in a stepwise fashion -ionization constant for each step -consider arsenic acid, H 3 AsO 4 which as three ionization constants K a1 = 2.5 x 10 -4 K a2 = 5.6 x 10 -8 K a3 = 3.0 x 10 -13 -first ionization step for arsenic acid H 3 AsO 4 ↔ H + + H 2 AsO - 4 K a1 = [H + ][H 2 AsO 4 - ]/[H 3 AsO 4 ] = 2.5 x 10 -4 H 2 AsO 4 1= 4 HAsO 4 2= ↔ H + + AsO 4 3- Salts NaCl – NaOH + HCl (does not hydrolyze because weaker than water)
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Unformatted text preview: NaCH 3 COO – NaOH + CH 3 COOH (will hydrolyze NH 4 Cl – NH 3 + HCl NH 4 CH 3 COO – NH 3 + CH 3 COOH The weaker the acid/base, the stronger the conjugate CH 3 COO-+ H 2 O ↔ CH 3 COOh + OH-*** NH 4 + + H 2 O ↔ NH 3 + H 3 O + Slide 66 ClO-+ H 2 O ↔ HClO + OH-Hydorolysis constant K b = [HClO][OH-]/[ClO-] Slide 69 KaKb = kw = 1x10-14 Slide 49 (Example 18-16-slide 50...
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