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Chemical Equilibrium: Additional Concepts

Chemical Equilibrium: Additional Concepts. Let’s see it again…. If Q sp < K sp , shift to products , no precipitate forms If Q sp > K sp , shift to reactants , precipitate will form If Q sp = K sp , no change will occur From your HW, #69.

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Chemical Equilibrium: Additional Concepts

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  1. Chemical Equilibrium: Additional Concepts Let’s see it again… • If Qsp < Ksp, shift to products, no precipitate forms • If Qsp> Ksp, shift to reactants, precipitate will form • If Qsp =Ksp, no change will occur • From your HW, #69

  2. Q 69) Will a precipitate form when 1.00 L of 0.150 M iron (II)chloride solution is mixed with 2.00 L of 0.0333 M sodium hydroxide solution? Ksp = 4.9x10-17 • Calculate Q • 1)Q expression FeCl2(aq) + 2NaOH(aq) Fe(OH)2(s) + 2NaCl(aq) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Qsp = [Fe2+] [OH-]2

  3. Q 69) Will a precipitate form when 1.00 L of 0.150 M iron (II)chloride solution is mixed with 2.00 L of 0.0333 M sodium hydroxide solution? Ksp = 4.9x10-17 • Calculate Q • 2) Molarities: [Fe2+] [OH-] .150 M .0333 M Ratio total volume: 1 L = 1 2 L = 2 3 L 3 3 L 3

  4. Q 69) Will a precipitate form when 1.00 L of 0.150 M iron (II)chloride solution is mixed with 2.00 L of 0.0333 M sodium hydroxide solution? Ksp = 4.9x10-17 • Calculate Q • 2) Molarities [OH-] = 2/3 (0.0333 M) = 0.0222 M [Fe2+] = 1/3 (0.150 M ) = 0.05 M Ratio of total volume

  5. Q 69) Will a precipitate form when 1.00 L of 0.150 M iron (II)chloride solution is mixed with 2.00 L of 0.0333 M sodium hydroxide solution? Ksp = 4.9x10-17 • Calculate Q • 3) Solve! • 4) Compare: Qsp = [Fe2+] [OH-]2 = [0.05M] [0.0222M]2 = 2.46x10-5 Qsp (2.46x10-5) Ksp(4.9x10-17) So… shift to reactants precipitate will form >

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