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Chapter 18: Chemical Equilibrium

Table of Contents. Chapter 18: Chemical Equilibrium. 18.3: Using equilibrium constants. Chemical Equilibrium: Basic Concepts. Equilibrium Expressions and Constants. What is K eq ?. Why do we use K eq ?. Write K eq for the following reaction:. Chemical Equilibrium: Additional Concepts.

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Chapter 18: Chemical Equilibrium

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  1. Table of Contents Chapter 18: Chemical Equilibrium 18.3: Using equilibrium constants

  2. Chemical Equilibrium: Basic Concepts Equilibrium Expressions and Constants • What is Keq? • Why do we use Keq? • Write Keq for the following reaction:

  3. Chemical Equilibrium: Additional Concepts Solubility equilibria • The solubility product constant(Ksp) is used for dissolving a sparingly soluble ionic compound in water. Equilibrium of a dissolved substance Tells us if we have more solid or more ions

  4. Chemical Equilibrium: Additional Concepts Solubility equilibria • When you know Ksp, you can… • 1. calculate the molar solubility of a sparingly soluble ionic compound • 2. calculate moles per liter of a saturated substance. • 3. calculate ion concentrations in a saturated solution. • These all use the same process!

  5. 1. Calculating Molar Solubility ex1)What is the molar solubility of copper II hydroxide? 1. Write the eq. expression 2. Look up Ksp 2.2 x 10-20 = [Cu2+][OH-]2 3. Replace Molarites with x 2.2 x 10-20 = [x][2x]2

  6. 1. Calculating Molar Solubility ex1)What is the molar solubility of copper II hydroxide? 2.2 x 10-20 = [x][2x]2 = x4x2 = 4x3 4. Solve for x [x]3 = 5.5 x 10-21 [x]= 1.8 x 10-7M The solubility of Cu(OH)2 is = 1.8 x 10-7 mol/L

  7. WE ALSO KNOW… The moles per liter of the saturated substance is 1.8x10-7mol/L x = [Cu2+]; 2x = [OH-] Ion concentrations [OH-] = 2x = 2(1.8 x 10-7 M) [Cu2+] = 1.8x10-7 M [OH-] = 3.6 x 10-7 M

  8. Practice Ex2) How many moles per liter of silver chloride will be in a saturated solution of AgCl? Ksp = 1.8x10-10 AgCl(s) Ag+(aq) + Cl-(aq) Ex3) Calculate the molar solubility of strontium chromate (SrCrO4)in water if Ksp = 3.7x10-5 1.3x10-5mol/L 1.3e-05 mol/L 0.00608 mol/L

  9. Chemical Equilibrium: Additional Concepts Predicting precipitates • The ion product (Q) • Ksp = Qsp at only at equilibrium • Q can predict the shift in equilibrium • Calculated the same way • Ksp = initial concentrations • Qsp = instantaneous concentrations

  10. Chemical Equilibrium: Additional Concepts Predicting precipitates • If Qsp < Ksp, shift to reactants, no precipitate forms • If Qsp> Ksp, shift to products, precipitate will form • If Qsp =Ksp, no change will occur

  11. Q ex4) Predict whether a precipitate of PbCl2 will form if 1L of 0.01M NaCl is added to 1L of 0.2M Pb(NO3)2. Ksp = 1.7x10-5 How did they know which precipitate? Solubility rules! Back of the periodic table

  12. Q ex4) Predict whether a precipitate of PbCl2 will form if 1L of 0.01MNaCl is added to 1L of 0.2MPb(NO3)2. Ksp = 1.7x10-5 • Calculate Q • 1)Q expression • 2) Molarities PbCl2(s) Pb2+(aq) + 2Cl-(aq) Qsp = [Pb2+] [Cl-]2 Total volume

  13. Q ex4) Predict whether a precipitate of PbCl2 will form if 1 L of 0.01 M NaCl is added to 1 L of 0.2 M Pb(NO3)2. Ksp = 1.7x10-5 • Calculate Q • 3) Solve! • 4) Compare: Qsp = [Pb2+] [Cl-]2 = [0.01M] [0.005M]2 = 2.5x10-7 Qsp (2.5x10-7) Ksp(1.7x10-5) So… shift to reactants no precipitate will form <

  14. Practice Ex5) Predict whether a precipitate of PbF2 will form when equal volumes of 0.10M Pb(NO3)2 and 0.30M NaF are mixed. Ksp = 3.3x10-8 PbF2(s) Pb2+(aq) + 2F-(aq) Ex6) Predict whether a precipitate Ag2SO4 will form when equal volumes of .25 M K2SO4 and .01 AgNO3 are mixed. Ksp = 1.2x10-5 Shift to reactants, precipitate of PbF2 Shift to products, no precipitate

  15. Chemical Equilibrium: Additional Concepts Common ion effect • Common ion effect: The solubility of a substance is reduced when the substance is dissolved in a solution containing a common ion. • For example, PbI2 is less soluble in an aqueous solution of NaI than in pure water. • Because the common ion I– is already present in the NaI solution. It reduces the maximum possible concentration of Pb2+ and thus reduces the solubility of PbI2.

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