2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect

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1 Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display The Common Ion Effect 17.2 Calculating the ph of a Buffer Preparing a Buffer Solution with as Specific ph 17.3 Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Strong Acid-Weak base Titrations Acid-Base Indicators 17.4 Solubility Equilibria Solubility Product Expression and K sp Calculations Involving K sp and Solubility Predicting Precipitation Reactions 17.5 The Common Ion Effect ph Complex Ion Formation 17.6 Separation of Ions Using Differences in Solubility Fractional Precipitation Qualitative Analysis of Metal Ions in Solution 17.1 The Common Ion Effect Worked Example 17.1 A system at equilibrium will shift in response to being stressed. The addition of a reactant or a product can be an applied stress. CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) Equilibrium concentration (M) 0.10 x x x Determine the ph at 25 C of a solution prepared by adding mole of sodium acetate to 1.0 L of 0.10-M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) Strategy Construct a new equilibrium table to solve for the hydrogen ion concentration. We use the stated concentration of acetic acid, 0.10 M, and [H + ] 0 M as the initial concentrations in the table. CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) [H + ] = [CH 3 COO ] = 1.34 x 10 3 M; ph = 2.87 CH 3 COONa(aq) CH 3 COOH(aq) H 2O Na + (aq) + CH 3 COO (aq) H + (aq) + CH 3 COO (aq) Equilibrium is driven toward reactant addition Equilibrium concentration (M) 0.10 x x x Worked Example 17.1 (cont.) Worked Example 17.1 (cont.) Solution These equilibrium concentrations are then substituted into the equilibrium expression to give (x)( x) = x Because we expect x to be very small (even smaller than M see above), because the ionization of CH 3 COOH is suppressed by the presence of CH 3 COO -, we assume (0.10 x) M 0.10 M and ( x) M M Therefore, the equilibrium expression simplifies to = (x)(0.050) and x = M. According to the equilibrium table, [H + ] = x, so ph = log( ) = Think About It The equilibrium concentrations of CH 3 COOH, CH 3 COO -, and H + are the same regardless of whether we add sodium acetate to a solution of acetic acid, add acetic acid to a solution of sodium acetate, or dissolve both species at the same time. We could have constructed an equilibrium table starting with the equilibrium concentrations in the 0.10 M acetic acid solution: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) Change in concentration (M) +y y y Equilibrium concentration (M) y y y In this case, the reaction proceeds to the left. (The acetic acid concentration increases, and the concentrations of hydrogen and acetate ions decrease.) Solving for y gives M. [H + ] = y = M and ph = We get the same ph either way. 1

2 17.2 A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer. acid conjugate base CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) reacts with added base CH 3 COOH(aq) + OH (aq) CH 3 COOH(aq) Buffer solutions resist changes in ph. reacts with added acid CH 3 COO (aq) + H 2 O(l) CH 3 COO (aq) + H + (aq) Calculate the ph of a buffer that contains 1.0 M acetic acid and 1.0 M sodium acetate. CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) Equilibrium concentration (M) 1.0 x x x x1.0 x 5 Ka x x = [H + ] = 1.8 x 10 5 M; ph = 4.74 Calculate the ph of the same buffer (1.0 M acetic acid and 1.0 M sodium acetate) after the addition of 0.10 mol of HCl. The added acid reacts with the conjugate base (acetate ion). Upon addition of H + : 1.0 mol 0.1 mol 1.0 mol CH 3 COO (aq) + H + (aq) CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) CH 3 COOH (aq) After H + has been consumed: 0.9 mol 0 mol 1.1 mol Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) CH 3 COOH(aq) x0.9 x 5 Ka x H + (aq) + CH 3 COO (aq) x +x +x 1.1 x x x x = [H + ] = 3.6 x 10 5 M; ph = 4.66 The original buffer had a ph = 4.74 Change in concentration (M) Equilibrium concentration (M) x +x +x 1.1 x x x The ph of a buffer solution can often be calculated with the Henderson-Hasselbalch equation. conjugate base ph p Ka log weak acid Worked Example 17.2 Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the ph after the addition of mole of NaOH. (Assume that the addition does not change the volume of the solution.) Strategy Added base will react with the acetic acid component of the buffer, converting OH - to CH 3 COO - : CH 3 COOH(aq) + OH - (aq) H 2 O(l) + CH 3 COO (aq) Write the starting amount of each species above the equation and the final amount of each species below the equation. Use the final amounts as concentrations in ph = pk a + log([a - ]/[HA]). Upon addition of OH - : 1.00 mol 0.10 mol 1.00 mol CH 3 COOH(aq) + OH - (aq) H 2 O(l) + CH 3 COO (aq) After OH - consumed: 0.90 mol 0 mol 1.10 mol 2

3 Worked Example 17.2 (cont.) Solution These equilibrium concentrations are then substituted into the equilibrium expression to give ph = log 1.10 M 0.90 M ph = 4.83 Thus, the ph of the buffer after addition of 0.10 mole of NaOH is Think About It Always do a reality check on a calculated ph. Although a buffer does minimize the effect of added base, the ph does increase. If you find that you ve calculated a lower ph after the addition of a base, check for errors like mixing up the weak acid and conjugate base concentrations or losing track of a minus sign. Buffers must have conjugate acid and base concentrations within a factor of 10. conjugate base weak acid The ph of a buffer cannot be more than one ph unit different than the pk a of the weak acid it contains. Worked Example 17.3 To make a buffer with a specific ph: 1) Pick a weak acid whose pk a is close to the desired ph. 2) Substitute the ph and pk a into the equation below to obtain the necessary [conjugate base]/[weak acid] ratio. A ph = pka log HA Select an appropriate weak acid from the table at right, and describe how you would prepare a buffer with a ph of Strategy Select an acid with a pk a within one ph unit of Use the pk a of the acid and ph = pk a + log([a - ]/[HA]) to calculate the necessary ratio of [conjugate base]/[weak acid]. Select concentrations of the buffer components that yield the calculated ratio. Weak Acid K a pk a HF HNO HCOOH C 6H 5COOH CH 3COOH HCN C 6H 5OH Solution Two of the acids listed have pk a values in the desired range: hydrocyanic acid (HCN, pk a = 9.31) and phenol (C 6 H 5 OH, pk a = 9.89) = log [C 6H 5 O - ] [C 6 H 5 OH] [C = log 6 H 5 O - ] = 0.39 [C 6 H 5 OH] [C 6 H 5 O - ] [C 6 H 5 OH] = = Worked Example 17.3 (cont.) 17.3 Solution Therefore, the ratio of [C 6 H 5 O - ] to [C 6 H 5 OH] must be 0.41 to 1. One way to achieve this would be to dissolve 0.41 mole of C 6 H 5 ONa and 1.00 mole of C 6 H 5 OH in 1.00 L of water. Think About It There is an infinite number of combinations of [conjugate base] and [weak acid] that will give the necessary ratio. Note that this ph could also be achieved using HCN and a cyanide salt. For most purposes, it is best to use the least toxic compounds available. Strong Acid-Strong Base Titrations The reaction between the strong acid HCl and the strong base NaOH can be represented by: NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) or by the net ionic equation, OH (aq) + H + (aq) H 2 O(l) 3

4 Titration of 25.0 ml of M HCl with M NaOH Titration of 25.0 ml of M HCl with M NaOH Strong Acid-Strong Base Titrations Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations: Weak Acid-Strong Base Titrations Consider the neutralization between acetic acid and sodium hydroxide: CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) The acetate ion that results from this neutralization undergoes hydrolysis: CH 3 COO (aq) + H 2 O(l)(aq) CH 3 COOH(aq) + OH (aq) The initial ph is determined by the ionization of acetic acid. CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial concentration (M) Equilibrium concentration (M) 0.10 x x x 2 x Ka x 5 x = [H + ] = 1.34 x 10 3 M; ph = 2.87 After the addition of base, some of the acetic acid has been converted to acetate ion: CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) Volume of OH added (ml) The solution is a buffer and the Henderson-Hasselbalch equation can be used to calculate ph. After 10.0 ml of base has been added: A ph = pka log 1.0 mmol ph = 4.74 log 4.56 HA 1.5 mmol OH added (mol) CH 3 COOH remaining CH 3 COO produced ph 4

5 At the equivalence point, all the acetic acid has been neutralized. CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) Volume of OH added (ml) OH added (mol) CH 3 COOH remaining CH 3 COO produced Must use TOTAL VOLUME to 2.5 mmol CH3COO M calculate concentration ml Initial concentration (M) Equilibrium concentration (M) x x x ph CH 3 COO (aq) + H 2 O(l) OH (aq) + CH 3 COOH (aq) CH 3 COO (aq) + H 2 O(l) OH (aq) + CH 3 COOH (aq) Initial concentration (M) Equilibrium concentration (M) x x x Use K a x K b = K w to get K b for the acetate ion; K b = 5.6 x CH COOH 2 OH x Kb CH COO x 3 10 x = [OH ] = 5.3 x 10 6 ; poh = 5.28; ph = 8.72 After the equivalence point, all the acetic acid has been neutralized, nothing is left to neutralize the added strong base. Weak Acid-Strong Base Titrations Volume of OH added (ml) OH added (mol) CH 3 COOH remaining CH 3 COO produced ph Volume of OH added (ml) OH added (mmol) Excess OH (mmol) Total Volume (ml) [OH ] (mol/l) poh ph Worked Example 17.4 Calculate the ph in the titration of 50.0 ml of M acetic acid by M sodium hydroxide after the addition of (a) 10.0 ml of base, (b) 25.0 ml of base, and (c) 35.0 ml of base. Strategy The reaction between acetic acid and sodium hydroxide is OH (aq) + CH 3 COOH (aq) CH 3 COO (aq) + H 2 O(l) Prior to the equivalence point [part (a)], the solution contains both acetic acid and acetate ion, making the solution a buffer. We can solve part (a) using the Henderson-Hasselbach equation. At the equivalence point [part (b)], all the acetic acid has been neutralized and we have only acetate ion in solution. We must determine the concentration of acetate ion and solve part (b) as an equilibrium problem, using the K b for acetate ion. After the equivalence point [part (c)], all the acetic acid has been neutralized and there is nothing to consume the additional added base. We must determine the concentration of excess hydroxide ion in the solution and solve for poh = -log[oh - ] and ph + poh = Worked Example 17.4 (cont.) Solution Remember that M can be defined as either mol/l or mmol/ml. For this type of problem, it simplifies the calculations to use millimoles rather than moles. K a for acetic acid is , so pk a = K b for acetate ion is (a) The solution originally contains (0.120 mmol/ml)(50.0 ml) = 6.00 mmol of acetic acid. A 10.0-mL amount of base contains (0.240 mmol/ml)(10.0 ml) = 2.40 mmol of base. After the addition of 10.0 ml of base, 2.40 mmol of OH - has neutralized 2.40 mmol of acetic acid, leaving 3.60 mmol of acetic acid and 2.40 mmol acetate ion in solution. Upon addition of OH - : 6.00 mol 2.40 mol 0 mol CH 3 COOH(aq) + OH - (aq) H 2 O(l) + CH 3 COO (aq) After OH - consumed: 3.60 mol 0 mol 2.40 mol ph = pka + log = =

6 Worked Example 17.4 (cont.) Worked Example 17.4 (cont.) Solution (b) After the addition of 25.0 ml of base, the titration is at the equivalent point. We calculate the ph using the concentration and the K b of acetate ion. At the equivalence point, we have 6.0 mmol of acetate ion in the total volume. We determine the total volume by calculating what volume of 0.24 M contains 6.0 mmol: (volume)(0.240 mmol/ml) = 6.00 mmol volume = 6.00 mmol mmol/ml = 25.0 ml Therefore, the equivalence point occurs when 25.0 ml of base has been added, making the total volume 50.0 ml ml = 75.0 ml. The concentration of acetate ion at the equivalence point is therefore 6.00 mmol CH 3 COO ml = M Solution (b) We can construct an equilibrium table using this concentration and solve for ph using the ionization constant for CH 3 COO - (K b = ): CH 3 COO - (aq) + H 2 O(l) OH - (aq) + CH 3 COOH(aq) Initial concentration (M) Equilibrium concentration (M) x x x Using the equilibrium expression and assuming that x is small enough to be neglected, K b = [CH 3COOH][OH - ] (x)(x) x 2 [CH 3 COO - = ] = x x = = M According to the equilibrium table, x = [OH - ], so [OH - ] = M. At equilibrium poh = log( ) = 5.17 and ph = = Titration of 25.0 ml of M NH 3 with M HCl Worked Example 17.5 Strong Acid-Weak Base Titrations Calculate the ph at the equivalence point when 25.0 ml of M NH 3 is titrated with M HCl. Strategy The reaction between NH 3 and HCl is NH 3 (aq) + H + (aq) NH 4+ (aq) At the equivalence point, all the NH 3 has been converted to NH 4+. Therefore, we must determine the concentration of NH 4 + at the equivalence point and use the K a for NH 4 + to solve for ph using an equilibrium table. Solution The solution originally contains (0.100 mmol/ml)(25.0 ml) = 2.50 mmol NH 4+. At the equivalence point, 2.50 mmol of HCl has been added. The volume of M HCl that contains 2.50 mmol is (volume)(0.100 mmol/ml) = 2.50 mmol volume = 2.50 mmol mmol/ml = 25.0 ml Worked Example 17.5 (cont.) Solution It takes 25.0 ml of titrant to reach the equivalence point, so the total solution is = 50.0 ml. At the equivalence point, all the NH 3 originally present has been converted to NH 4+. The concentration of NH + 4 is (2.50 mmol)/(50.0 ml) = M. We must use this concentration as the starting concentration of ammonium ion in our equilibrium table. Think About It In the titration NH 4+ (aq) of a weak + H 2 O(l) base with NH a strong 3 (aq) acid, + H 3 O + (aq) the species in solution at the equivalence point is the conjugate acid. Initial Therefore, concentration we (M) should expect an acidic ph. Once all the NH 0 3 has 0 been converted to NH 4+, there is no longer anything in the solution Change to consume in concentration the added (M) acid. Thus, x the ph after the equivalence +x point +x depends on the number of millimoles of H + added and not consumed Equilibrium divided concentration by the new total (M) volume x x x The equivalence point in a titration can be visualized with the use of an acid-base indicator. HIn(aq) The endpoint of a titration is the point at which the color of the indicator changes. H + (aq) + In (aq) The equilibrium expression is K a = [NH 3][H + ] (x)(x) x 2 = [NH 4+ ] = x x = = M = [H + ] ph = log( ) =

7 Worked Example 17.6 Which indicator listed in Table 17.3 would you use for the acid-base titrations shown in (a) Figure 17.3, (b) Figure 17.4, and (c) Figure 17.5? Strategy Determine the ph range that corresponds to the steepest part of each titration curve and select an indicator (or indicators) that changes color within that range. Solution (a) The titration curve at right is for the titration of a strong acid with a strong base. The steep part of the curve spans a ph range of about 4 to 10. Most of the indicators in Table 17.3, with the exceptions of thymol blue, bromophenol blue, and methyl orange, would work for the titration of a strong acid with a strong base. Worked Example 17.6 (cont.) Solution (b) The figure at right shows the titration of a weak acid with a strong base. The steep part of the curve spans a ph range of about 7 to 10. Cresol red and phenolphthalein are suitable indicators. Think About It If we don t select an appropriate indicator, the endpoint (color change) will not coincide with the equivalence point. (c) The figure at right shows the titration of a weak base with a strong acid. The steep part of the curve spans a ph range of 7 to 3. Bromophenol blue, methyl orange, methyl red, and chlorophenol blue are all suitable indicators Solubility Equilibria Solubility Equilibria Quantitative predictions about how much of a given ionic compound will dissolve in water is possible with the solubility product constant, K sp. AgCl(s) Ag + (aq) + Cl (aq) K sp = [Ag + ][Cl ] Compound Dissolution Equilibrium K sp Aluminum hydroxide Al(OH) 3 (s) Al 3+ (aq) + 3OH (aq) 1.8 x Calcium fluoride CaF 2 (s) Ca 2+ (aq) + 2F (aq) 4.0 x Silver bromide AgBr(s) Ag + (aq) + Br (aq) 7.7 x Silver chloride AgCl(s) Ag + (aq) + Cl (aq) 1.6 x 10 6 Molar solubility is the number of moles of solute in 1 L of a saturated solution (mol/l) Solubility is the number of grams of solute in 1 L of a saturated solution (g/l). To calculate a compound s molar solubility: 1) Construct an equilibrium table. 2) Fill in what is known. 3) Determine the unknowns. Zinc sulfide ZnS(s) Zn 2+ (aq) + S 2 (aq) 3.0 x

8 Solubility Equilibria The K sp of silver bromide is 7.7 x Calculate the molar solubility. AgBr(s) Ag + (aq) + Br (aq) Initial concentration (M) 0 0 Change in concentration (M) +s +s Equilibrium concentration (M) s s K sp = [Ag + ][Br ] = 7.7 x x = (s)(s) s = 8.8 x 10 7 M mol AgBr g 1 L 1 mol AgBr g/l Worked Example 17.7 Calculate the solubility of copper(ii) hydroxide [Cu(OH) 2 ] in g/l. Strategy Write the dissociation equation for Cu(OH) 2, and look up its K sp value in Table Solve for molar solubility using the equilibrium expression. Convert molar solubility to solubility in g/l using the molar mass of Cu(OH) 2. Solution The equation for the dissociation of Cu(OH) 2 is Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) and the equilibrium expression is K sp = [Cu 2+ ][OH - ] 2. According to Table 17.4, K sp for Cu(OH) 2 is The molar mass of Cu(OH) 2 is g/mol. Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) Initial concentration (M) 0 0 Change in concentration (M) +s +2s Worked Example 17.7 (cont.) Solution Therefore, s = = (s)(2s) 2 = 4s = M The molar solubility of Cu(OH) 2 is M. Multiplying by its molar mass gives solubility of Cu(OH) 2 = -7 mol g Cu(OH) 2 1 mol Cu(OH) 2 = L -5 g/l Think About It Common errors arise in this type of problem when students neglect to raise an entire term to the appropriate power. For example, (2s) 2 is equal to 4s 2 (not 2s 2 ). Equilibrium concentration (M) s 2s Worked Example 17.8 The solubility of calcium sulfate (CaSO 4 ) is measured experimentally and found to be 0.67 g/l. Calculate the value of K sp for calcium sulfate. Strategy Convert solubility to molar solubility using the molar mass of CaSO 4, and substitute the molar solubility into the equilibrium expression to determine K sp. Think About It The K sp for CaSO 4 is relatively large (compared to Solution many The of the molar K sp mass values of in CaSO Table 4 is 17.4) In g/mol. fact, sulfates The molar are solubility listed as of CaSO 4 soluble is compounds in Table molar solubility of CaSO 4 = , g but CaSO calcium sulfate 4 1 L 1 mol is CaSO listed as an insoluble exception. Remember that the term insoluble really 4 refers to compounds that are slightly soluble, and that different g CaSO sources 4 may differ with regard to s how = soluble -3 mol/l a compound must be to be The equation considered and the soluble. equilibrium expression for the dissociation of CaSO 4 are CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) and K sp = [Ca 2+ ][SO 4 2- ] substituting the molar solubility into the equilibrium expression gives K sp = (s)(s) = ( ) 2 = Solubility Equilibria For the dissociation of an ionic solid in water, the following conditions may exist: 1) The solution is unsaturated 2) The solution is saturated 3) The solution is supersaturated The following relationships are useful in making predictions on when a precipitate might form. Q < K sp ; no precipitate forms Q = K sp ; no precipitate forms Q > K sp ; a precipitate forms 8

9 Worked Example 17.9 Worked Example 17.9 (cont.) Predict whether a precipitate will form when each of the following is added to 650 ml of M K 2 SO 4 : (a) 250 ml of M BaCl 2 ; (b) 175 ml of 0.15 M AgNO 3 ; (c) 325 ml of 0.25 M Sr(NO 3 ) 2. (Assume volumes are additive.) Strategy For each part, identify the compound that might precipitate and look up its K sp value in Table 17.4 or Appendix 4. Determine the concentration of each compound s constituent ions, and use them to determine the value of the reaction quotient, Q sp ; then compare each reaction quotient with the value of the corresponding K sp. If the reaction quotient is greater than K sp, a precipitate will form. Solution (a) BaSO 4 might form and its K sp = Concentrations of the constituent ions of BaSO 4 are: [Ba 2+ ] = 250 ml M = M [SO 4 2- ] = 650 ml M = M Using these 650 concentrations ml ml in the equilibrium expression, 650 ml [Ba ][SO ml 2-4 ], gives a reaction quotient of (0.0011)(0.0058) = , which is greater than the K sp of BaSO 4 ( ). Therefore, BaSO 4 will precipitate. Solution (b) Ag 2 SO 4 might form and its K sp = Concentrations of the constituent ions of Ag 2 SO 4 are: [Ag ml 0.15 M ] = = M [SO ml ml ml 4 ] = = M M Using these Think concentrations About It Students in the equilibrium sometimes have expression, difficulty 650 ml [Ag deciding ml + ] 2 [SO 2- what 4 ], gives a reaction compound quotient of might (0.032) precipitate. 2 (0.0063) Begin = by writing -6, which down is less the constituent than the K sp of Ag 2 SO ions 4 ( in the two -5 ). Therefore, solutions before Ag 2 SO they 4 will are not combined. precipitate. Consider the two possible combinations; the cation from the first solution and the (c) SrSO anion 4 might from form the second, and its K or vice sp = versa. -7 You. Concentrations can consult the of the constituent ions of information Ag 2 SO 4 are: in Table 9.2 and 9.3 to determine whether one of the combinations is insoluble. Also keep in mind that only an insoluble salt will have a tabulated K [Sr 2+ sp value. ] = = M [SO 2-4 ] = = M 325 ml 0.25 M 650 ml ml 650 ml M 650 ml ml Using these concentrations in the equilibrium expression, [Sr 2+ ][SO 2-4 ], gives a reaction quotient of (0.083)(0.0053) = , which is greater than the K sp of SrSO 4 ( ). Therefore, SrSO 4 will precipitate Worked Example Several factors exist that affect the solubility of ionic compounds: Common ion effect ph Formation of complex ions Calculate the molar solubility of silver chloride in a solution that is M in silver nitrate. Strategy Silver nitrate is a strong electrolyte that dissociates completely in water. Therefore, the concentration of Ag + before any AgCl dissolves is M. Use the equilibrium expression, the K sp for AgCl, and an equilibrium table to determine how much AgCl will dissolve. Solution The dissolution equilibrium and the equilibrium expression are AgCl(s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] AgCl(s) Ag + (aq) + Cl - (aq) Initial concentration (M) Change in concentration (M) +s +s Equilibrium concentration (M) s s Worked Example (cont.) Solution Substituting these concentrations into the equilibrium expression gives = ( s)(s) We expect s to be very small, so s and = ( )(s) Thus s = = M Therefore, the molar solubility of AgCl -3 in M AgNO 3 is M. 10 Think About It The molar solubility of AgCl in water is = M. The presence of M AgNO3 reduces the solubility of AgCl by a factor of ~500. ph: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH (aq) K sp = [Mg 2+ ][OH ] 2 = 1.2 x (s)(2s) 2 = 4s 3 = 1.2 x s = 1.4 x 10 4 M At equilibrium: [OH ] = 2(1.4 x 10 4 M ) = 2.8 x 10 4 M poh = log(2.8 x 10 4 ) = 3.55 ph = = In a solution with a ph of less than 10.45, the solubility of Mg(OH) 2 increases. 9

10 ph: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH (aq) ph: BaF 2 (s) Ba 2+ (aq) + 2F (aq) 2H + (aq) + 2OH (aq) 2H 2 O(l) 2H + (aq) + 2F (aq) 2HF(aq) Overall: Mg(OH) 2 (s) + 2H + (aq) Mg 2+ (aq) + 2H 2 O(l) Overall: BaF 2 (s) + 2H + (aq) Ba 2+ (aq) + 2HF(aq) If the ph of the medium were higher than 10.45, [OH ] would be higher and the solubility of Mg(OH) 2 would decrease because of the common ion (OH - ) effect. As the concentration of F decreases, the concentration of Ba 2+ must increase so satisfy the equality: K sp = [Ba 2+ ][F ] 2 The solubilities of salts containing anions that do not hydrolyze are unaffected by ph: Cl, Br, NO 3 Worked Example Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO 4? Strategy For each salt, write the dissociation equilibrium equation and determine whether it produces an anion that will react with H +. Only an anion that is the conjugate base of a weak acid will react with H +. Solution The dissolution equilibrium and the equilibrium expression are (a) CuS(s) Cu 2+ (aq) + S 2- (aq) S 2- is the conjugate base of the weak acid HS -. S 2- reacts with H + as follows: S 2- (aq) + H + (aq) HS - (aq) Worked Example (cont.) Solution CuS and PbSO 4 are more soluble in acid than in water. (AgCl is no more or less soluble than in water.) Think About It When a salt dissociates to give the conjugate base of a weak acid, H + ions in an acidic solution consume a product (base) of the dissolution. This drives the equilibrium to the right (more solid dissolves) according to Le Châtelier s principle. (b) AgCl(s) Ag + (aq) + Cl - (aq) Cl - is the conjugate base of the strong acid HCl. Cl - does not react with H +. (c) PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) SO 4 2- is the conjugate base of the weak acid HSO 4-. It reacts with H + as follows: SO 4 2- (aq) + H + (aq) HSO 4- (aq) Complex ion formation: A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. A solution of CoCl 2 is pink because of the presence of Co(H 2 O) 6 2+ ions. Complex ion formation: When HCl is added to a CoCl 2 solution, there is a color change from pink to blue. This solution is blue because of the presence of presence of CoCl 4 2 ions. Co 2+ (aq) + 4Cl (aq) CoCl 4 2 (aq) 10

11 Complex ion formation: Cu 2+ (aq) + 2OH (aq) Cu(OH) 2 (s) Cu(OH) 2 (s) + 4NH 3 (aq) Cu(NH 3 ) 4 2+ (aq) + 2OH (aq) The formation of the Cu(NH 3 ) 4 2+ ion can be expressed as Cu 2+ (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 2+ (aq) 2 2 NH3 Cu NH Kf Cu Worked Example In the presence of aqueous cyanide, cadmium(ii) forms the complex ion Cd(CN) Determine the molar concentration of free (uncomplexed) cadmium(ii) ion in solution when 0.20 mol of Cd(NO 3 ) 2 is dissolved in a liter of 2.0 M sodium cyanide (NaCN). Strategy Because formation constants are typically very large, we begin by assuming that all the Cd 2+ ion is consumed and converted to complex ion. We then determine how much Cd 2+ is produced by the subsequent dissociation of the complex ion, a process for which the equilibrium constant is the reciprocal K f. Solution From Table 17.5, the formation constant (K f ) for the complex ion Cd(CN) 4 2- is The reverse process, Cd(CN) 4 2- (aq) Cd 2+ (aq) + 4CN - (aq) has an equilibrium constant of 1/K f = The equilibrium expression for the dissociation is = [Cd2+ ][CN - ] 4 [Cd(CN) 4 2- ] Worked Example (cont.) 17.6 Separation of Ions Using Differences in Solubility Solution The formation of complex ion will consume some of the cyanide originally present. Stoichiometry indicates that four CN - ions are required to react with one Cd 2+ ion. Therefore, the concentration of CN - that we enter in the top row of the equilibrium table will be [2.0 M 4(0.20 M)] = 1.2 M. Cd(CN) 2-4 (aq) Cd 2+ (aq) + 4CN - (aq) Initial concentration (M) Think About It When you assume 0.20 that all the metal 0 ion is 1.2 consumed and converted to complex ion, Change in concentration (M) -x it s important +x to remember +4x that some of the complexing agent (in this case, CN - ion) is consumed Equilibrium in the concentration process. Don t (M) forget 0.20 to - adjust x its concentration x x accordingly before entering it in the top row of the equilibrium table. and, because the magnitude of K is so small, we can neglect x with respect to the initial concentrations of Cd(CN) 2-4 and CN - (0.20 x 0.20 and x 1.2), so the solution becomes [Cd 2+ ][CN - ] 4 x(1.2) 4 = [Cd(CN) 2- = - 4 ] and x = M. Some compounds can be separated based on fractional precipitation. Fractional precipitation is the separation of mixture based upon the components solubilities. Compound K sp AgCl 1.6 x AgBr 7.7 x AgI 8.3 x

12 Worked Example Worked Example (cont.) Silver nitrate is added slowly to a solution that is M in Cl- ions and M in Br - ions. Calculate the concentration of Ag + ions (in mol/l) required to initiate the precipitation of AgBr without precipitating AgCl. Strategy Silver nitrate dissociates in solution to give Ag + and NO 3 - ions. Adding Ag + ions in sufficient amount will cause the slightly soluble ionic compounds AgCl and AgBr to precipitate from solution. Knowing the K sp values for AgCl and AgBr (and the concentrations of Cl - and Br - already in solution), we can use the equilibrium expressions to calculate the maximum concentration of Ag + that can exist in solution without exceeding K sp for each compound. Solution The solubility equilibria, K sp values, and equilibrium expressions for AgCl and AgBr are AgCl(s) Ag + (aq) + Cl - (aq) K sp = = [Ag + ][Cl - ] AgBr(s) Ag + (aq) + Br - (aq) K sp = = [Ag + ][Br - ] Because the K sp for AgBr is smaller (by a factor of more than 200), AgBr should precipitate first; that is, it will require a lower concentration of added Ag + to begin precipitation. Therefore, we first solve for [Ag + ] using the equilibrium expression for AgBr to determine the minimum Ag + concentration necessary to initiate precipitation of AgBr. We then solve for [Ag + ] again, using the equilibrium expression for AgCl to determine the maximum Ag + concentration that can exist in the solution without initiating precipitation of AgCl. Solving the AgBr equilibrium expression for Ag + concentration, we have K [Ag + ] = sp [Br - ] = - 13 = M Worked Example (cont.) Solution For AgBr to precipitate from solution, the silver ion concentration must exceed M. Solving the AgCl equilibrium expression for the Ag + concentration, we have K [Ag + sp ] = [Cl - = - ] 10 = M For AgCl not to precipitate from solution, the silver ion concentration must stay below M. Therefore, to precipitate the Br - ions without precipitating the Cl - from this solution, the Ag + concentration must be greater than M and less than M. Think About It If we continue adding AgNO 3 until the Ag + concentration is high enough to begin precipitation of AgCl, the concentration of Br - remaining in solution can also be determined using the K sp expression. Separation of Ions Using Differences in Solubility Qualitative analysis involves the principle of selective precipitation and can be used to identify the types of ions present in a solution. K [Br - sp ] = [Ag + = - ] 13 = Thus, by the time AgCl begins to precipitate, ( M -5 M) (0.020 M) = , 9 so less than 0.5 percent of the original bromide ion remains in the solution. 17 Key Concepts Calculating the ph of a Buffer Preparing a Buffer Solution with as Specific ph Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Strong Acid-Weak base Titrations Acid-Base Indicators Solubility Product Expression and K sp Calculations Involving K sp and Solubility Predicting Precipitation Reactions The Common Ion Effect ph Complex Ion Formation Fractional Precipitation Qualitative Analysis of Metal Ions in Solution 12