# Updates Assignment 06 is due Mon., March 12 (in class) Midterm 2 is Thurs., March 15 and will cover Chapters 16 & 17 –Huggins 10, 7-8pm –For conflicts:

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Updates Assignment 06 is due Mon., March 12 (in class) Midterm 2 is Thurs., March 15 and will cover Chapters 16 & 17 –Huggins 10, 7-8pm –For conflicts: ELL 221, 6-7pm (must arrange at least one week in advance)

Acid-Base Equilibria and Solubility Equilibria Chapter 17

Titration of a Weak Acid with a Strong Base Conjugate base of the weak acid affects the pH when it is formed. The pH at the equivalence point will be >7.

Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

Titration of a Weak Acid with a Strong Base Four regions where you might need to calculate pH Look to next slide for helpful hints

Titration of a Weak Acid with a Strong Base The initial pH: pH equals the pH of the weak acid solution (use initial concentration and K a to solve for x, then calculate pH) Between initial pH and equivalence point: to determine pH in this range, we must consider neutralization of the acid (calculate new amounts of weak acid and its conjugate base, then use pK a and the Henderson-Hasselbalch equations to calculate pH) Equivalence point: to determine pH at this point, we must consider the amount of weak base formed in the complete neutralization of weak acid with strong base because the weak base generates hydroxide ions in a reaction with water (use concentration of weak base and K b to solve for x and report as pOH, use the “pH + pOH = 14” relationship to calculate pH After the equivalence point: In this region, hydroxide from reaction of the conjugate base with water is negligible compared to the hydroxide from excess NaOH (calculation is same as that for titrating a strong acid with a strong base, assume all of the weak acid is neutralized and any remaining hydroxide from NaOH is used in the pH calculation assuming complete ionization of NaOH) CH 3 CO 2 H (aq) + OH − (aq)  CH 3 CO 2 − (aq) + H 2 O (l)

Titration of a Weak Acid with a Strong Base Note that the titration curves for the titrations of both strong acid and weak acid are the same after the equivalence point.

Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

Solubility equilibria Equilibria considered thus far have involved acids and bases and have been homogeneous Now we will consider the equilibria involved in the dissolution or precipitation of ionic compounds which are classified as heterogeneous equilibria

Heterogeneous equilibria Tooth enamel dissolves in acidic solutions, causing tooth decay Precipitation of certain salts in the kidneys produces kidney stones Precipitation of CaCO 3 from groundwater is responsible for the formation of stalactites and stalagmites within limestone caves

Heterogeneous equilibria Earlier you learned solubility rules (see pg. 97) to give a qualitative sense of whether a compound will have a low or a high solubility in water By considering solubility equilibria, we can make more quantitative predictions

Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2− ] where the equilibrium constant, K sp, is called the solubility product.

Solubility Products K sp is not the same as solubility. Solubility of a substance is the quantity that dissolves to form a saturated solution Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L), or in mol/L (M). K sp is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution and is a unitless number. Its magnitude is a measure of how much of the solid dissolves to form a saturated solution.

Solubility Products Solubility of a substance can change considerably as the concentrations of other solutes change, i.e., the solubility of Mg(OH) 2 depends highly on pH and the presence of other sources of Mg 2+ In contrast, K sp has only one value for a given solute at any specfic temperature (holds strictly only for dilute solutions, high ionic strengths cause deviations but we will ignore these)

Solubility Products Solubility of a any compound in g/L can be converted to molar solubility The molar solubility can be used to determine the concentration of ions in solution The concentrations of ions can be used to calculate K sp The steps can be reversed, and solubility can be calculated from K sp

Solubility Equilibria 17.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 3 3 - ] 2

17.6

What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10 17.6

Precipitation and separation of ions Predicting whether a precipitate will form Predicting what precipitate will from from a mixture of ions Predicting a selective precipitation of an ion from a mixture of ions

Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid will dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp. Remember Q is the reaction quotient, which is obtained by substituting the initial concentrations into the equilibrium expression.

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? 17.6 The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form

What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 17.7 AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.020 = = 3.9 x 10 -11 M [Ag + ] = K sp [Cl - ] 1.6 x 10 -10 0.020 = = 8.0 x 10 -9 M 3.9 x 10 -11 M < [Ag + ] < 8.0 x 10 -9 M

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