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Copyright McGraw-Hill 20091 Chapter 17 Acid-Base Equilibria and Solubility Equilibria Insert picture from First page of chapter

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Copyright McGraw-Hill 20092 17.1 The Common Ion Effect When a compound containing an ion in common with an already dissolved substance is added to a solution at equilibrium, the equilibrium shifts to the left. This phenomenon is known as the common ion effect. Produced by the addition of a second solute.

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Copyright McGraw-Hill 20093 Example: 1.0 L of 0.10 M solution of CH 3 COOH Add 0.050 mol of CH 3 COONa Effect

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Copyright McGraw-Hill 20094 Effect on equilibrium calculations Without the common ion 0.10 M 0.10 M − x − x+x x x

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Copyright McGraw-Hill 20095 Effect on equilibrium calculations With the common ion 0.10 M 0.10 M − x − x+ x x 0.050 M + x 0.050 M

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Copyright McGraw-Hill 20096 17.2 Buffer Solutions Buffer –Solution containing a weak acid and its conjugate base –Solution containing a weak base and its conjugate acid –Resist changes in pH on addition of small amount of acid on addition of small amount of base –Calculation of the pH of a buffer pH is a common ion problem

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Copyright McGraw-Hill 20097 Henderson-Hasselbalch equation Quantitative equation based on the K a expression Expressed using -logs

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Copyright McGraw-Hill 20098 0.100 L of solution

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Copyright McGraw-Hill 20099 Add 0.001 mol of HCl, a strong acid. Before reaction: 0.001 mol 0.010 mol 0.010 mol After reaction: 0 mol 0.009 mol 0.011 mol 0.100 L of solution

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Copyright McGraw-Hill 200910 Before reaction: 0.001 mol 0.010 mol 0.010 mol After reaction: 0 mol 0.009 mol 0.011 mol Add 0.001 mol of NaOH, a strong base. 0.100 L of solution

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Copyright McGraw-Hill 200911 Without a buffer, the pH changes drastically. water Add 0.001 mol H + Add 0.001 mol OH -

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Copyright McGraw-Hill 200912 Preparing a Buffer with a Specific pH –Concentration condition –Effective range

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Copyright McGraw-Hill 200913 Select an appropriate acid, and describe how you would prepare a buffer with pH 4.5.

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Copyright McGraw-Hill 200914 possible acids Using C 6 H 5 COOH

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Copyright McGraw-Hill 200915 Dissolve 2.04 mol of C 6 H 5 COONa and 1.00 mol of C 6 H 5 COOH in enough water to form 1.00 L of solution. If the solubility of the substances does not permit these amounts to dissolve, reduce the amounts but maintain the same ratio. Dissolve 0.408 mol of C 6 H 5 COONa and 0.20 mol of C 6 H 5 COOH in enough water to form 1.00 L of solution.

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Copyright McGraw-Hill 200916 17.3 Acid-Base Titrations Titration – addition of a solution of accurately known concentration to a solution of unknown concentration until the reaction is complete. –Standard solution - one of known concentration –Equivalence point – the point when stochiometrically equivalent amounts have been added –Endpoint – the point in the laboratory when the titration is stopped –Titrant – the solution that is place in the buret

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Copyright McGraw-Hill 200917 Types of titration systems to be considered –Strong acid – strong base –Weak acid – strong base –Strong acid – weak base

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Copyright McGraw-Hill 200918 Typical points to measure pH during a titration –Initial pH –Between initial pH and the equivalence point –Equivalence point –After the equivalence point Titration curve – a graph of pH as a function of volume of titrant added

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Copyright McGraw-Hill 200919 Strong acid –strong base titration Net ionic equation pH at points in the titration is determined by stoichiometry and the excess reagent. Titration of 0.100 M NaOH (buret) and 25.0 mL 0.100 M HCl (pH being monitored)

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Copyright McGraw-Hill 200928 Weak acid –strong base titration –More complex equations govern pH due to equilibrium reactions –pH at points in the titration are determined by a specific equilibrium reaction. –Titration of 0.100 M NaOH (buret) and 25.0 mL 0.100 M CH 3 COOH (pH being monitored) Overall reaction At equivalence point

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Copyright McGraw-Hill 200936 Identical to strong acid- strong base titration

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Copyright McGraw-Hill 200938 Strong acid –weak base titration –Similar calculations to Weak acid-strong base titration –pH at points in the titration are determined by a specific equilibrium reaction. –Titration of 0.100 M HCl (buret) and 25.0 mL 0.100 M NH 3 (pH being monitored) Overall reaction At equivalence point

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Copyright McGraw-Hill 200939 Initial pH

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Copyright McGraw-Hill 200941 Equivalence point pH

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Copyright McGraw-Hill 200943 Acid-base indicator –A weak organic acid or base for which the ionized and un-ionized forms are different colors. –pH range over which the indicator changes color depends on the magnitude of the K a or K b –Use to signal endpoints during a titration –The pH at the equivalence point must be within the pH range where the indicator changes color.

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Copyright McGraw-Hill 200946 An Indicator Derived from Red Cabbage pH increases

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Copyright McGraw-Hill 200947 17.4 Solubility Equilibria Solubility Product Expression and K sp –K sp is the solubility product constant –Set up like other equilibrium expression –General example; MX n (s) ⇄ M n+ (aq) + n X (aq) –Solids and liquids are not included in equilibrium expressions

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Copyright McGraw-Hill 200949 Molar solubility (s)– number of moles of solute in 1 L of a saturated solution (mol/L) Solubility - number of grams of solute in 1 L of a saturated solution (g/L) K sp can be used to determine molar solubility (and solubility) –Handle as an equilibrium problem –Use an equilibrium table Molar solubility can be used to determine the value of the K sp.

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Copyright McGraw-Hill 200950 Calculate the solubility of SnS in g/L at 25°C.

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Copyright McGraw-Hill 200951 s +s s

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Copyright McGraw-Hill 200952 The solubility of lead(II) chromate (PbCrO 4 ) is 4.5 x 10 5 g/L. Calculate the solubility product (K sp ) of lead(II) chromate.

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Copyright McGraw-Hill 200954 Precipitation can be predicted –Compare the reaction quotient (Q) to the K sp where “i” designates the initial concentration –Q < K sp no precipitate will form –Q > K sp precipitate will form until Q = K sp MX n (s) ⇄ M n+ (aq) + n X (aq)

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Copyright McGraw-Hill 200955 17.5 Factors Affecting Solubility The Common Effect – an example of LeChâtelier’s principle –The presence of a second salt that produces an ion common to a solubility equilibrium will reduce solubility. Example: AgCl in a solution of AgNO 3 –The concentration of a product ion is increased forcing the solubility reaction toward reactant, the solid. Example: Ag + from the AgNO 3, reverses the solubility reaction.

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Copyright McGraw-Hill 200956 Calculate the molar solubility of BaSO 4 in 0.0010 M Na 2 SO 4.

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Copyright McGraw-Hill 200957 ss 0.0010 M 0.0010 M + s +s s

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Copyright McGraw-Hill 200958 Key Points The Common Ion Effect Buffer Solutions –pH of a buffer –Preparation of a buffer Acid-Base Titrations –Strong acid-strong base –Weak acid-strong base –Strong acid-weak base

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Copyright McGraw-Hill 200959 –Acid-base indicators Solubility Equilibria –Solubility product expression –K sp –Calculation of K sp –Calculation of solubility –Predicting precipitation reactions Comparison of Q and K sp Factors Affecting Solubility –Common ion effect

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Copyright McGraw-Hill 200960 pH Complex ion formation –Complex ions –Formation constants Separation of Ions Using Solubility Differences –Fractional precipitation –Qualitative analysis of metal ions

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