Lesson_20_-_Example_Solutions-2009

# Lesson_20_-_Example_Solutions-2009 - 2Ag CrO 4 2-K sp = 1.9...

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Lesson 20 Example 20-2 Calculate the mass of AgCl that can dissolve in 100 mL of 0.15 M NaCl solution. Data: K sp (AgCl) = 1.6x10 -10 mol 2 /L 2 Solution: Basis: 1.0 L of solution Data: M AgCl = 143.3 g/mol K sp = [Ag + ][Cl - ] = 1.6 × 10 -10 mol 2 /L 2 In equilibrium: [Cl - ] = (0.15 + S ) [Ag - ] = S 1

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Lesson 20 Therefore, S (0.15 + S ) = 1.6 × 10 -10 mol 2 /L 2 S 2 + 0.15 S – 1.6 × 10 -10 = 0 ( 29 ( 29 ( 29 2 10 x 6 . 1 1 4 15 . 0 15 . 0 S 10 2 - + ± - = S 1 = 1.0665x10 -9 mol/L or S 2 = -0.15 mol/L (no physical meaning) ( 29 = - AgCl mol AgCl g 3 . 143 mL 100 mL 1000 L 1 L AgCl mol 10 x 07 . 1 m 9 AgCl = 1.53x10 -8 g If no NaCl is present in solution: K sp = [Ag + ][Cl - ] = S 2 S = ( K sp ) 1/2 = (1.6 × 10 10 ) 1/2 = 1.26 × 10 -5 mol/L or 1.81 × 10 -4 g in 100 mL 2
Lesson 20 Example 20-3 Consider a sparingly soluble compound with stoichiometry Ag 2 (CrO 4 )(s) Ag 2 (CrO 4 ) (s)

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Unformatted text preview: 2Ag + + CrO 4 2-K sp = 1.9 × 10-12 = [Ag + ] 2 [CrO 4 2-] Let S = molar solubility of Ag 2 CrO 4 (s) = (2S) 2 (S) = 4S 3 Consider the solubility of Ag 2 CrO 4 (s) in an aqueous solution of 0.01 M Ag NO 3 Note Ag NO 3 ⇌ Ag + + NO 3-complete dissociation is valid. Ag 2 CrO4(s) ⇌ 2 Ag + (aq) + CrO 4 2-(aq) Init 3 5 3 sp 10 8 . 7 4 K S-× = = Lesson 20 0.01 M Ag NO 3 Equilibrium (0.01 + 2 S) (S) K sp = (0.01 + 2S) 2 (S) = 1.9 × 10-12 Assume 2 S << 0.01 The presence of 0.01 M AgNO 3 has decreased the solubility of Ag 2 CrO 4 by more than 3 orders of magnitude and the assumption that 2 S << 0.01 is valid. 4 8 4 12 10 9 . 1 10 10 9 . 1 S---× = × = ∴...
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Lesson_20_-_Example_Solutions-2009 - 2Ag CrO 4 2-K sp = 1.9...

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