Solubility Product Constant Chapter 7.6
Types of Equilibrium: I. Chemical equilibrium: dynamic equilibrium between reactants and products in a closed system H2(g) + I2(g) ↔ 2 HI(g) II. Phase equilibrium: a dynamic equilibrium between different physical states of the same pure substance in a closed system H2O(l) ↔ H2O(g)
Types of Equilibrium: III. Solubility equilibrium: dynamic equilibrium between a solute and solvent in a saturated solution in a closed system CO2(g) ↔CO2(aq) • What are saturated Solutions? Ex: Barium sulfate (BaSO4) is an ionic solid that has certain solubility in H2O(l). If enough BaSO4 is added to a solution of H2O(l) the solution will become saturated, and some solid BaSO4 will remain.
Solubility Equilibrium • In the case of the saturated solution, an equilibrium is established between the solid and dissolved ions of BaSO4 BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) • Is it an example of homogenous or heterogeneous equilibrium? • In writing the equilibrium expression we omit the "concentration" of the solid component(s)
SOLUBILITY PRODUCT (Ksp) • Ksp is the solubility product constant (or just the solubility product) • Note that although the concentration for the solid component(s) is omitted, for the equilibrium to exist some un-dissolved BaSO4must be present • In other words, Ksp = [Ba2+(aq)][SO42-(aq)] is true only when some solid BaSO4 is present in the solution • Ksp describes the equilibrium concentrations of dissolved ions that exists under conditions of saturation with the solid form
SOLUBILITY PRODUCT (Ksp) • The smaller the value of Ksp, the lower the solubility of the ions of an ionic solid • Arrange the following substances in order of increasing solubility: cobalt (II) hydroxide calcium phosphate lead(II) chloride lithium carbonate (refer to page 484 of textbook for Ksp) • Writing the expression of Ksp is similar for other equilibrium constants; • it is equal to the product of the concentrations of the dissolved ions raised to the power of the coefficients from the balanced equation
SOLUBILITY PRODUCT (Ksp) Sample Problem #1: Write equilibrium expressions for the solubility of CaF2 and Ca3(PO4)2. CaF2(s) Ca+2 + 2F-1 Ksp= [Ca+2][F-1]2 Ca3(PO4)2(s) 3Ca+2 + 2PO4-3 Ksp= [Ca+2]3 [PO4-3]2
Background: Predicting ions from formulas Q - Write the ions that form from Al2(SO4)3(aq)? Step 1 - look at the formula: Al2(SO4)3(aq) Step 2 - determine valences: Al3 (SO4)2 • (subscripts become the charges or look at the group numbers - Al is 3+ according to the periodic table) • Step 3 - write ions: 2Al3+(aq) + 3SO42–(aq) • Note that there are 2 aluminums because Al has a subscript of 2 in the original formula
Practice with writing ions Q - Write ions for Na2CO3(aq) A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+. There are 2, thus we have 2Na+. There is only one CO3. It must have a 2- charge) • Notice that when ions form from molecules, charge can be separated, but the total charge (and number of each atom) stays constant. Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq) A - 3Ca2+(aq) + 2PO43–(aq) A - Cd2+(aq) + 2NO3–(aq) Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq) A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)
How do we find the concentration of the [ions] from the concentration of chemical given to us? • How many moles of K+ are present in 2 L of 1.5 M KCl? • How many moles of K+ are present in 1 L of 3 M K2SO4? • What is the concentration (in mol/L) of K+ when 2 L of 1.5 M KCl is mixed with 1 L of 3 M K2SO4? (Hint: Add together the # mol K+ from each source to get total # mol K+. Divide this by total number of L.) • How many grams of K+ are present in Q#1 & 2?
How to calculate Ksp? • Sample Problem #2: If 0.0067g of CaCO3 is dissolved in 1.0L of water, calculate its Ksp. 100g of CaCO3 → 1 mole of CaCO3 0.0067g of CaCO3 → (1/100)*0.0067 moles So molar solubility of CaCO3 = 6.7x10-5 M CaCO3(s) ↔ Ca+2 + CO3-26.7x10-5M 6.7x10-5M 6.7x10-5M Ksp = [Ca+2][CO3-2] = [6.71x10-5][6.7x10-5] Ksp = 4.5x10-9
How to calculate the molar solubility if Ksp is given? • If Ksp of an ionic compound is known at 25°C, its molar solubility (concentration of solute that dissolved) can be determined using ICE tables. • Refer to page #485-486 of the textbook. Look at the sample problem #2. • Practice solving Q#1 on page#486 of the textbook.
HOMEWORK • Practice the following questions: • Q#4 on page 486 of the textbook. • Q# 7, 8, 12, 13 on page 493 of the textbook (one of these questions will be on test/exam)
Precipitation • Precipitation refers to the formation of a solid from ions. A precipitate is “insoluble” and as a result it settles at the bottom of the reaction mixture. • It usually forms as a result of a double displacement reaction: AB + CD AD + CB • ions switch their places: AB(aq) + CD(aq) AD(s) + CB(aq) Recall: aq indicates the compound is aqueous (as ions)
Solubility • Refers to the concentration of a saturated solution of a solute in a particular solvent at a particular temperature. • Solubleand insoluble are general terms to describe how much of a solid dissolves. • Solubility can be predicted from rules (pg.801) • These are general rules, based on observation • Todeterminesolubility,followthechart • You will not have to memorize these rules, you will have to use the rules to predict solubility • Use the chart to predict the solubility of the following compounds: Ca(NO3)2 FeCl2 Ni(OH)2 AgNO3 BaSO4 CuCO3
Solubility CHART – 801 a) Ca(NO3)2 - Soluble (salts containing NO3- are soluble) b) FeCl2 - Soluble (all chlorides are soluble) c) Ni(OH)2 - Insoluble (all hydroxides are insoluble) d) AgNO3 - Soluble (salts containing NO3- are soluble) e) BaSO4 - Insoluble (Sulfates are soluble, except … Ba2+) f) CuCO3 - Insoluble (containing CO32- are insoluble)