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Chapter 9 Chemical Equilibrium

Chapter 9 Chemical Equilibrium. 9.6 Equilibrium in Saturated Solutions. Saturated Solution. A saturated solution contains the maximum amount of dissolved solute contains solid solute is an equilibrium system: rate of dissolving = rate of recrystallization

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Chapter 9 Chemical Equilibrium

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  1. Chapter 9 Chemical Equilibrium 9.6Equilibrium in Saturated Solutions

  2. Saturated Solution A saturated solution • contains the maximum amount of dissolved solute • contains solid solute • is an equilibrium system: rate of dissolving = rate of recrystallization solid ions in solution

  3. Solubility Product Constant The solubility product constant for a saturated solution • gives the ion concentrations at constant temperature • is expressed as Ksp • does not include the solid, which is constant Fe(OH)2(s) Fe2+(aq) + 2OH−(aq) Ksp = [Fe2+] [OH−]2

  4. Learning Check Write the Kspexpression for each of the following: A. FeS(s) B. Ag2CO3(s) C. Ca(IO3)2(s)

  5. Solution Write the Kspexpression for each of the following: A. FeS(s) Fe2+(aq) + S2−(aq) Ksp = [Fe2+][S2−] B. Ag2CO3(s) 2Ag+(aq) + CO32− (aq) Ksp = [Ag+]2 [CO32−] C. Ca(IO3)2(s) Ca2+(aq) + 2IO3−(aq) Ksp = [Ca2+][IO3−]2

  6. Guide to Calculating Ksp

  7. Example of Calculating Solubility Product Constant Calculate the Kspof PbSO4 (solubility 1.4 x 10–4 M). STEP 1 Write the equilibrium equation for dissociation: PbSO4(s) Pb2+(aq) + SO42−(aq) STEP 2 Write the Kspexpression: Ksp = [Pb2+][SO42−] STEP 3 Substitue molarity values and calculate: Ksp = (1.4 x 10–4) x (1.4 x 10–4) = 2.0 x 10–8

  8. Examples of Solubility Product Constants

  9. Learning Check What is the Ksp value of PbF2 if the solubility at 25 C is 2.6 x 10–3 M? 1) 1.4 x 10–5 2) 6.8 x 10–6 3) 7.0 x 10–8

  10. Solution 3) 7.0 x 10-8 STEP 1 Write the equilibrium equation for dissociation: PbF2(s) Pb2+(aq) + 2F−(aq) STEP 2 Write the Kspexpression: Ksp = [Pb2+][F−]2 STEP 3 Substitute molarity values and calculate: [Pb2+] = 2.6 x 10–3 M [F−] = 2 x [Pb2+] = 5.2 x 10–3 M Ksp = (2.6 x 10–3) x (5.2 x 10–3)2 = 7.0 x 10–8

  11. Molar Solubility (S) The molar solubility (S) is • the number of moles of solute that dissolve in 1 L of solution • determined from the formula of the salt • calculated from the Ksp

  12. Calculating Molar Solubility (S)

  13. Solubility Calculation Determine the solubility (S)2 of SrCO3 (Ksp = 5.4 x 10–10). STEP 1 Write the equilibrium equation for dissociation: SrCO3(s) Sr2+(aq) + CO32−(aq) STEP 2 Write the Kspexpression: Ksp = [Sr2+][CO32−] STEP 3 Substitute S for the molarity of each ion into Ksp: Ksp = [Sr2+][CO32−] = [S][S] = S2 = 5.4 x 10−10 STEP 4 Calculate the solubility, S: S = = 2.3 x 10−5 M

  14. Learning Check Calculate the solubility (S) of PbSO4 , if theKsp = 1.6 x 10–8. 1) 1.3 x 10–4 M 2) 4.0 x 10–4 M 3) 2.6 x 10–16 M

  15. Solution 1) 1.3 x 10–4 M STEP 1 Write the equilibrium equation for dissociation: PbSO4(s) Pb2+(aq) + SO42−(aq) STEP 2 Write the Kspexpression: Ksp = [Pb2+][SO42−] STEP 3 Substitute S for the molarity of each ion into Ksp: Ksp = [Pb2+][SO42−] = [S][S] = S2 = 1.6 x 10–8 STEP 4 Calculate the solubility, S: S = = 1.3 x 10–4 M

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