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CHEMISTRY 1220
CHAPTER 17 PRACTICE EXAM
All questions listed below are problems taken from old Chemistry 123 exams given here at The Ohio State University. Read Chapter 17.4 – 17.7 and complete the following problems. These problems will not be graded or collected, but will be very similar to what you will see on an exam.
Ksp and Molar Solubility (Section 17.4)
The solubility product expression equals the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium expression. For example: PbCl 2 (s) Pb2+(aq) + 2Cl-‐(aq) Ksp=[Pb2+] [Cl-‐]^2 Why isn’t the concentration of PbCl 2 in the denominator of the Ksp? Recall from Section 15.4 that whenever a pure solid or pure liquid is in a heterogeneous (multiple phases present) equilibrium, its concentration is not included in the equilibrium expression.
1. The solubility product expression for La 2 (CO 3 ) 3 is Ksp =? First, write the equilibrium expression. The subscript following each atom in the solid is the number of ions of that atom in solution. The total charges must add to zero: La 2 (CO 3 ) 3 (s) 2La3+(aq) + 3CO 32 -‐(aq) Given the information above and the equilibrium expression: Ksp=[La3+]^2 [CO 32 -‐]^3 2. The solubility product expression for Zn 3 (PO 4 ) 2 is Ksp =? Zn 3 (PO 4 ) 2 (s) 3Zn2+(aq) + 2PO 43 -‐(aq) Ksp=[Zn2+]^3 [PO 43 -‐]^2 3. The solubility product expression for Al 2 (CO 3 ) 3 is Ksp =? Al 2 (CO 3 ) 3 (s) 2 Al3+(aq) + 3CO 32 -‐(aq) Ksp=[Al3+]^2 [CO 32 -‐]^3 4. The solubility product expression for Fe(OH) 3 is Ksp =? Fe(OH) 3 (s) Fe3+(aq) + 3OH-‐(aq) Ksp=[Fe3+][OH-‐]^3 Solubility differs from the solubility product constant in that solubility is the amount of a substance that dissolves to form a saturated solution. It is often expressed as grams of solute per liter of solution or as number of moles of solute per liter solution (molar solubility). It can change if the concentrations of other ions or pH change. Ksp is a unit-‐less number that is the equilibrium constant for the solid-‐ion equilibrium. It is a measure of how much the solid dissolves to give a saturated
solution. It is constant for a given solute at a specific temperature. However, because both deal with saturated solutions, Ksp can be calculated from solubility and vice versa. 5. Calculate the molar solubility of CrF 3 in water. Ksp for CrF 3 is 6.6 x 10-‐^11. To calculate solubility from Ksp, set up an ICE table for the equilibrium expression. The column for the solid can be ignored because its concentration does not appear in the equilibrium expression. Assume initially that none of the salt has dissolved (initial concentrations= 0 M). The change in concentration will be +nx M, where n=the coefficient of the ion in the equilibrium expression. Equilibrium concentrations are the resultant concentrations from the “initial” and “change” rows (i.e. the sum): CrF 3 (s) _____Cr3+(aq) + 3F-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 0 M C hange -‐-‐-‐-‐-‐ +1x M +3x M E quilibrium -‐-‐-‐-‐-‐ +x M +3x M These equilibrium concentrations can be substituted into the solubility product expression. Since Ksp is given, we can solve for x, which is the molar solubility: Ksp=[Cr3+] [F-‐]^3 = (x)(3x)^3 = 27x^4 = 6.6 x 10-‐^11 x= 1.3 x 10 -‐^3 M 6. What is the molar solubility for PbCrO 4? Ksp PbCrO 4 = 2.8 x 10-‐^13 PbCrO 4 (s) ____Pb^2 +(aq) + CrO 42 -‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 0 M C hange -‐-‐-‐-‐-‐ +1x M +1x M E quilibrium -‐-‐-‐-‐-‐ +x M +x M Ksp=[Pb^2 +] [CrO 42 -‐]= (x)(x)= x^2 = 2.8 x 10-‐^13 x= 5 .3 x 10 -‐^7 M 7. The solubility of BaF 2 is 1.3 g/L. What is the solubility product constant? Conversely, Ksp can be calculated from either solubility or molar solubility. First, convert solubility in g/L to mol/L (M) since the concentrations used in Ksp are in mol/L (M). If the given solubility is already in mol/L, this step is unnecessary: 1.3 g BaF 2 _ x 1 mol BaF 2 __ = 0.0074 mol/L= 0.0074 M BaF 2 1 L 175.33 g BaF 2 From this molar concentration and the equilibrium expression, we can calculate the equilibrium concentrations of each ion using their respective coefficients in the equilibrium: BaF 2 (s) Ba^2 +(aq) + 2F-‐(aq) 0.0074 M BaF 2 x 1 mol Ba2+^ = 0.0074 M Ba2+ 1 mol BaF 2
Be careful! This question asks for the solubility of iodide ions , not necessarily that of the ionic compound. PbI 2 (s) ____Pb^2 +(aq) + 2I-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 0 M C hange -‐-‐-‐-‐-‐ +1x M +2x M E quilibrium -‐-‐-‐-‐-‐ +x M +2x M Ksp=[ Pb^2 +] [I-‐]^2 = (x)(2x)^2 = 4x^3 = 1.4 x 10-‐^8 x= 1.5 x 10-‐^3 M HOWEVER, this is NOT the final answer because, according to the ICE table, the equilibrium concentration (solubility) of iodide ions in the saturated solution is 2 x: 2x=2(1.5 x 10-‐^3 M)= 3.0 x 10 -‐^3 M 16. Which compound is the least soluble (mol/L) in water? (a) CaCO 3 Ksp = 2.8 x 10–^9 (b) PbI 2 Ksp = 8.7 x 10–^9 (c) AgBr Ksp = 5.0 x 10–^13 (d) Fe(OH) 2 Ksp = 8.0 x 10–^16 (e) Co(OH) 2 Ksp = 1.6 x 10–^15 You can directly compare solids with the same ratio of cation to anion (i.e. higher Ksp, more soluble) to eliminate some possible answer choices. However, if the ratios differ, the molar solubility (x) for EACH compound must be calculated using an ICE table because the lowest Ksp does not necessarily correspond to the lowest molar solubility. (a) 5.29 x 10-‐^5 mol/L (b) 1.30 x 10-‐^3 mol/L (c) 7.07 x 10-‐^7 mol/L (d) 5.85 x 10-‐^6 mol/L (e) 7.37 x 10-‐^6 mol/L Of these values, (c) is the lowest. So AgBr is the least soluble (even though its Ksp is not the lowest). 17. Which of the following salts has the highest concentration of silver ion when dissolved in water? AgCl Ksp = 1.6 x 10-‐^10 Ag 2 CO 3 Ksp = 8.1 x 10-‐^12 AgBr Ksp = 5.0 x 10-‐^13 Calculate the molar concentration of silver ion for each of the salts. (a) 1.26 x 10-‐^5 M (b) 2.53 x 10 -‐^4 M (make sure to multiply x by 2!) (c) 7.07 x 10-‐^7 M Of these concentrations, (b) is the highest. So Ag 2 CO 3 has the highest concentration of silver ion (Note: this is not just because silver has a subscript of 2 after it in Ag 2 CO 3 ; you must calculate the actual concentration for each compound). 18. Which compound is least soluble in water?
a) Co(OH) 2 Ksp = 1.6 x 10-‐^15 b) Fe(OH) 2 Ksp = 8.0 x 10-‐^16 c) AgBr Ksp = 5.0 x 10-‐^13 d) CaCO 3 Ksp = 2.8 x 10-‐^9 e) PbI 2 Ksp = 8.7 x 10-‐^9 Make sure to calculate molar solubility and not to merely compare Ksp values if the ion ratios diffrer. (a) 7.37 x 10-‐^6 mol/L (b) 5.85 x 10-‐^6 mol/L (c) 7.07 x 10-‐^7 mol/L (d) 5.29 x 10-‐^5 mol/L (e) 1.30 x 10-‐^3 mol/L Of these values, (c) is the lowest. So AgBr is the least soluble.
Qualitative Analysis Group I (Section 17.7 & “Isolation and Characterization of Metal Ions:
Exploitation of Differences in Solubility” Lab)
19. An aqueous solution contains the following twelve ions: Zn2+, Co2+, Pb2+, Ni2+, Cu2+, Sn2+, Fe2+, Sb3+, Al3+, Cr3+, Bi3+, and Ag+^ and cold dilute HCl is added. What precipitate(s) will form? Look at the solubility rules. All chloride salts are soluble EXCEPT those with Ag+, Hg2+, and Pb2+. Therefore, of the cations listed, PbCl 2 (s) and AgCl(s) will form. 20. Why must the dilute HCl added in to the solution in question 1 9 be cold? PbCl 2 is slightly soluble in water. Solubility decreases as temperature decreases. So by using cold water, we ensure that PbCl 2 (s) precipitates from the solution. 21. Why is it better to use dilute HCl in question 1 9 dilute rather than use concentrated HCl? We use dilute HCl to prevent the precipitation of basic chlorides and hydroxides. 22. In the group I analysis, what is the basis of separation for AgCl(s) and PbCl 2 (s). Why is this separation possible? By heating the solution, we separate PbCl 2 and AgCl because the solubility of PbCl 2 increases dramatically with increasing temperature, so its precipitate dissolves upon heating. However, the solubility of AgCl remains fairly constant, so it remains solid. 23. What color is PbCrO 4 (s)? Yellow
Factors Influencing Solubility (Section 17.5)
2 4. How many moles of MgF 2 (Ksp = 6.4 x 10–^9 ) will dissolve in 0.50 L of 0.20 M NaF? This is an example of the common ion effect. Because some F-‐^ is already in the solution due to the presence NaF, the solubility of MgF 2 will decrease because its solubility equilibrium will shift left away from the “added ion”, in accordance with LeChatelier’s Principle: MgF 2 (s) Mg2+(aq) + 2F-‐(aq) To find the number of moles of MgF 2 that will dissolve in 0.50 L of solution, we must calculate the molar solubility of MgF 2 in this solution, using an ICE table. **Remember, because NaF is completely
28. As the pH decreases, how will the solubility of Cu(OH) 2 be affected? To determine the effect of an acid/base or change in pH, write out the solubility equilibrium expression: Cu(OH) 2 (s) Cu2+(aq) + 2OH-‐(aq) If the pH is decreased, the concentration of H+^ ions increases. These H+^ ions can neutralize the OH-‐ ions from the above equilibrium, forming H 2 O, and this decreases the concentration of OH-‐^ in solution. According to LeChatelier’s Principle, the equilibrium will then shift to the right , thereby **increasing the solubility of Cu(OH) 2 (s).
- The Ksp for Zn(OH) 2 is 5.0** × 10 -‐^17. Determine the molar solubility of Zn(OH) 2 in buffered solution with a pH of 11.50? First, determine the concentration of OH-‐^ when the pH is 11.50: pH=-‐log[H+]=11. [H+]=3.16 x 10-‐^12 M [H+][OH-‐]=1.0 x 10-‐^14 (always true) (3.16 x 10-‐^12 M) +][OH-‐]=1.0 x 10-‐^14 [OH-‐]=3.16 x 10-‐^3 M Now, the problem is similar to a common-‐ion problem. However, because the solution is buffered, it resists changes in pH. Therefore, the concentration of OH-‐^ does NOT change : Zn(OH) 2 (s) ___ Zn^2 +(aq) + 2OH-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 3.16 x 10-‐^3 M C hange -‐-‐-‐-‐-‐ +1x M NO CHANGE E quilibrium -‐-‐-‐-‐-‐ x M 3.16 x 10-‐^3 M Ksp=[ Zn^2 +] [OH-‐]^2 =(x)( 3.16 x 10-‐^3 )^2 =5.0 x 10-‐^17 x= 5 .0 x 10 -‐^12 M 30. Calculate the solubility of Cu(OH) 2 in a solution buffered at pH = 8.50. Ksp for Cu(OH) 2 = 1.6 x 10-‐^19 pH=-‐log[H+]=8. [H+]=3.16 x 10-‐^9 M [H+][OH-‐]=1.0 x 10-‐^14 (always true) (3.16 x 10-‐^9 M) +][OH-‐]=1.0 x 10-‐^14 [OH-‐]=3.16 x 10-‐^6 M Cu(OH) 2 (s) ___ Cu^2 +(aq) + 2OH-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 3.16 x 10-‐^6 M C hange -‐-‐-‐-‐-‐ +1x M NO CHANGE E quilibrium -‐-‐-‐-‐-‐ x M 3.16 x 10-‐^6 M Ksp=[ Cu^2 +] [OH-‐]^2 =(x)( 3.16 x 10-‐^6 )^2 =1.6 x 10-‐^19 x= 1.6 x 10 -‐^8 M
31. Calculate the molar solubility of Mn(OH) 2 when buffered at pH = 11.40. The Ksp for Mn(OH) 2 is 1.6 x 10-‐^13. pH=-‐log[H+]=11. [H+]=3.98 x 10-‐^12 M [H+][OH-‐]=1.0 x 10-‐^14 (always true) (3.98 x 10-‐^12 M) +][OH-‐]=1.0 x 10-‐^14 [OH-‐]=2.51 x 10-‐^3 M Mn(OH) 2 (s) _ Mn^2 +(aq) + 2OH-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 2.51 x 10-‐^3 M C hange -‐-‐-‐-‐-‐ +1x M NO CHANGE E quilibrium -‐-‐-‐-‐-‐ x M 2.51 x 10-‐^3 M Ksp=[ Mn^2 +] [OH-‐]^2 =(x)( 2.51 x 10-‐^3 )^2 =1.6 x 10-‐^13 x= 2.5 x 10 -‐^8 M 32. What is the pH of a saturated solution of Cu(OH) 2 (Ksp = 2.6 × 10 -‐^19 )? To determine the pH, we need to determine the concentration of H+^ using an ICE table: Cu(OH) 2 (s) ___ Cu^2 +(aq) + 2OH-‐(aq) I nitial -‐-‐-‐-‐-‐ 0 M 0 M C hange -‐-‐-‐-‐-‐ +1x M +2x M E quilibrium -‐-‐-‐-‐-‐ x M 2x M Ksp=[Cu^2 +][OH-‐]^2 =(x)( 2x)^2 =4x^3 =2.6 x 10-‐^19 x= 4.0 x 10-‐^7 M Remember to multiply x by 2 in this problem because the equilibrium concentration of OH-‐^ is 2 x: [OH-‐]= 2 x=2(4.0 x 10-‐^7 M)= 8.0 x 10-‐^7 M [H+][OH-‐]=1.0 x 10-‐^14 [H+](8.0 x 10-‐^7 M )=1.0 x 10-‐^14 [H+]=1.2 x 10-‐^8 M pH=-‐log[H+]=-‐log(1.2 x 10-‐^8 )=7. 33. What is the molar solubility (mol/L) of Cr(OH) 3 at pH = 10.00? Ksp for Cr(OH) 3 is 6.3 x 10–^31 and Kf for Cr(OH) 4 – is 8 x 10^29.* In this problem, a complex ion is formed. This effect increases solubility and trumps both the common-‐ion and pH effects. To determine the molar solubility of a solid whose ion forms a complex ion, we need to find the overall equilibrium expression and “overall K” for both the dissolution of the solid and the formation of the complex-‐ion. Identical species on opposite sides of the equilibrium arrow cancel. When “adding” two equilibria, the overall K is the product of the individual K values: Cr(OH) 3 (s) Cr3+(aq) + 3OH-‐(aq) Ksp=6.3 x 10-‐^31 Cr3+(aq) + 4 OH-‐(aq) Cr(OH 4 )-‐(aq) Kf=8.0 x 10^29 Cr(OH) 3 (s) + OH-‐(aq) Cr(OH 4 )-‐(aq) K= Ksp x Kf =(6.3 x 10-‐^31 )(8.0 x 10^29 ) =5.04 x 10-‐^1 Calculate the initial concentration of OH-‐^ from the given pH: pH=-‐log[H+]=10.
35. Calculate the molar solubility of CdCO 3 in 1.5 M NH 3. Note that Cd2+^ forms the Cd(NH 3 ) 4 2+^ complex ion for which Kf is 1.3 x 10^7. Ksp for CdCO 3 is 5.2 x 10-‐^12. CdCO 3 (s) Cd^2 +(aq) + CO 32 -‐(aq) Ksp=5.2 x 10-‐^12 Cd^2 +(aq) + 4 NH 3 (aq) Cd(NH 3 ) 4 2+(aq) Kf=1.3 x 10^7 CdCO 3 (s) + 4NH 3 (aq) Cd(NH 3 ) 4 2+(aq)+ CO 32 -‐(aq) K= Ksp x Kf =(5.2 x 10-‐^12 )(1.3 x 10^7 ) =6.76 x 10-‐^5 CdCO 3 (s) + 4NH 3 (aq) Cd(NH 3 ) 4 2+(aq) + CO 32 -‐(aq) I nitial -‐-‐-‐-‐-‐-‐ 1.5 M 0 M 0 M C hange -‐-‐-‐-‐-‐-‐ -‐4x M +x M +x M E quilibrium -‐-‐-‐-‐-‐-‐ 1.5-‐4x M x M x M K=[Cd(NH 3 ) 4 2+][CO 32 -‐]= x^2 ____ = 6.76 x 10-‐^5 [NH 3 ]^4 (1.5-‐4x)^4 x= 1.8 x 10 -‐^2 M 36. What is the concentration of free cobalt ion in a solution that is 4.75 x 10 -‐^2 M Co2+^ and 5.0 M NH 3? Kf Co(NH 3 ) 6 2+ = 8.3 x 10 4 Co2+(aq) + 6NH 3 (aq) Co(NH 3 ) 6 2+(aq) I nitial 4.75 x 10-‐^2 M 5.0 M 0 M C hange -‐4.75 x 10-‐^2 M -‐6(4.75 x 10-‐^2 ) M +4.75 x 10-‐^2 M F inal (NOT equilibrium) 0 M 4.715 M 4.75 x 10-‐^2 M Co(NH 3 ) 6 2+(aq) Co2+(aq) + 6NH 3 (aq) I nitial 4.75 x 10-‐^2 M 0 M 4.715 M C hange -‐x M +x M +6x M E quilibrium 4.75 x 10-‐^2 -‐x M x M 4.715+6x M Kd=[Co2+][NH 3 ]^6 = (x)(4.715+ 6 x)^6 = (1.1 x 10^4 )(x)= 1 _____ [Co(NH 3 ) 6 2+] (4.75 x 10-‐^2 -‐x) 4.75 x 10-‐^2 8.3 x 10^4 x= 5.2 x 10 -‐^11 M 37. Calculate the molar solubility of AgCl in 12 M NH 3. Ksp AgCl = 1.8 x 10-‐^10 Kf Ag(NH 3 ) 2 = 1.7 x 10^7 AgCl(s) Ag+(aq) + Cl-‐(aq) Ksp=1.8 x 10-‐^10 Ag+(aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 +(aq) Kf=1.7 x 10^7 AgCl(s) + 2NH 3 (aq) Ag(NH 3 ) 2 +^ + Cl-‐(aq) K= Ksp x Kf =3.06 x 10-‐^3 AgCl(s) + 2NH 3 (aq) Ag(NH 3 ) 2 +^ + Cl-‐(aq) I nitial -‐-‐-‐-‐-‐-‐ 12 M 0 M 0 M C hange -‐-‐-‐-‐-‐-‐ -‐2x M +x M +x M E quilibrium -‐-‐-‐-‐-‐-‐ 12 -‐2x M x M x M K=[Ag(NH 3 ) 2 +][Cl-‐]= ___x^2 ____ = 3.06 x 10-‐^3 [NH 3 ]^2 (12-‐ 2 x)^2 x= 6.6 x 10 -‐^1 M
38. Calculate the concentration of free aluminum ion, [Al3+], in 1.0 L of solution that contains 0.040 mol Al(NO 3 ) 3 and 2.00 mol NaF. Kf for AlF 63 -‐^ is 7.1 x 10^19. Al3+(aq) + 6F-‐(aq) AlF 63 -‐(aq) I nitial 0.040 M 2.00 M 0 M C hange -‐0.040 M -‐6(0.040) M +0.040 M F inal (NOT equilibrium) 0 M 1.76 M 0.040 M AlF 63 -‐^ (aq) Al3+(aq) + 6F-‐(aq) I nitial 0.040 M 0 M 1.76 M C hange -‐x M +x M +6x M E quilibrium 0.040-‐x M x M 1.76+6x M Kd=[Al^3 +][F-‐]^6 = (x)(1.76+ 6 x)^6 = (2.97 x 10^1 )(x)=___ 1 _____ [AlF 63 -‐] (0.040 -‐x) 0.040 7.1 x 10^19 x= 1.9 x 10 -‐^23 M 39. The Cd2+^ ion forms the complex ion CdCl 42 -‐^ for which Kf = 6.3 x 10^2. Determine the equilibrium constant for the solubility of CdCO 3 in contact with a solution that contains Cl-‐^ ion. Ksp for CdCO 3 is 5.2 x 10-‐^12. CdCO 3 (s) Cd^2 +(aq) + CO 32 -‐(aq) Ksp=5.2 x 10-‐^12 Cd^2 +(aq) + 4 Cl-‐(aq) CdCl 42 -‐^ (aq) Kf=6.3 x 10^2 CdCO 3 (s) + 4Cl-‐(aq) CdCl 42 -‐(aq)+ CO 32 -‐(aq) K= Ksp x Kf =(5.2 x 10-‐^12 )(6.3 x 10^2 ) K=3.28 x 10 -‐^9 40. Use information from problem 39 to calculate the molar solubility of CdCO 3 in 3.0 M NaCl. CdCO 3 (s) + 4Cl-‐(aq) CdCl 42 -‐(aq) + CO 32 -‐(aq) I nitial -‐-‐-‐-‐-‐-‐ 3.0 M 0 M 0 M C hange -‐-‐-‐-‐-‐-‐ -‐4x M +x M +x M E quilibrium -‐-‐-‐-‐-‐-‐ 3.0-‐4x M x M x M K=[CdCl 4 2+][CO 32 -‐]= ____x^2 ____ = 3.28 x 10-‐^9 [Cl-‐]^4 (3.0-‐4x)^4 x= 5.2 x 10 -‐^4 M 41. The Ag+^ ion forms the complex ion AgCl 2 -‐^ for which Kf = 2.5 x 10^5. Determine the equilibrium constant for the solubility of AgCl in the presence of excess chloride ion. Ksp of AgCl is 1.8 x 10-‐^10. AgCl(s) Ag+(aq) + Cl-‐(aq) Ksp=1.8 x 10-‐^10 Ag+(aq) + 2 Cl-‐(aq) AgCl 2 -‐(aq) Kf=2.5 x 10^5 AgCl(s) + Cl-‐(aq) AgCl 2 -‐^ K= Ksp x Kf K=4.5 x 10 -‐^5 42. Use the information from problem 41 to calculate the molar solubility of AgCl in 8.5 M HCl. AgCl(s) + Cl-‐(aq) AgCl 2 -‐(aq) I nitial -‐-‐-‐-‐-‐-‐ 8.5 M 0 M C hange -‐-‐-‐-‐-‐-‐ -‐x M +x M E quilibrium -‐-‐-‐-‐-‐-‐ 8.5-‐x M x M
46. The formation constant, Kf, for Ni(NH 3 ) 6 2+^ is 5.5 × 108. What is the concentration of free nickel ions in a solution that contains 0.045 M Ni2+^ and 3.0 M NH 3 (concentrations refer to the moment before the formation of the complex ion)? Ni2+(aq) + 6NH 3 (aq) Ni(NH 3 ) 6 2+(aq) I nitial 0.045 M 3.0 M 0 M C hange -‐0.045 M -‐6(0.045) M +0.045 M F inal (NOT equilibrium) 0 M 2.73 M 0.045 M Ni(NH 3 ) 6 2+(aq) Ni2+(aq) + 6NH 3 (aq) I nitial 0.045 M 0 M 2.73 M C hange -‐x M +x M +6x M E quilibrium 0.045 -‐x M x M 2.73 +6x M Kd=[Ni2+][NH 3 ]^6 = (x)( 2.73 + 6 x)^6 = (4.1 x 10^2 )(x)=___ 1 _____ [Ni(NH 3 ) 6 2+] (0.045 -‐x) 0.045 5.5 x 10^8 x= 2.0 x 10 -‐^13 M 47. A solution is saturated with silver acetate, AgC 2 H 3 O 2 (Ksp = 1.9 x 10–^3 ). Which of the following reagents will increase the solubility of silver acetate? NaC 2 H 3 O 2 HNO 3 NH 3 AgNO 3 For problems like these, first write out the solubility equilibrium expression and see how the various reagents will influence the equilibrium. Remember, a common-‐ion will decrease solubility, the effect of a change in pH will depend on the particular equilibrium, and formation of a complex ion will increase solubility, trumping both the common-‐ion and pH effects: AgC 2 H 3 O 2 (s) Ag+(aq) + C 2 H 3 O 2 -‐(aq) NaC 2 H 3 O 2 -‐common ion, therefore decreases solubility. HNO 3 -‐ C 2 H 3 O 2 -‐^ is the conjugate base of a weak acid (HC 2 H 3 O 2 ), and is therefore a weak base itself. It can be neutralized by HNO 3 , a strong acid, thereby decreasing the concentration of C 2 H 3 O 2 -‐^ and shifting the equilibrium to the right, increasing solubility. NH 3 -‐ Ag+^ forms the complex ion Ag(NH 3 ) 2 +, increasing solubility. AgNO 3 -‐ common ion, therefore decreases solubility. 48. How many of the following salts would be more soluble in acidic solution that in pure water? BaC 2 O 4 CaS AuCl 3 PbF 2 ZnCO 3 For an acidic solution to increase solubility, it must neutralize a weak base. Therefore the anion must be the conjugate base of a weak acid. Remember, conjugate salts of strong acids/bases are NOT influenced by addition of acids/bases or pH changes: BaC 2 O 4 -‐ C 2 O 4 -‐^ is the conjugate base of the weak acid HC 2 O 4. Solubility increases. CaS-‐ S^2 -‐^ is the conjugate base of the weak acid HS-‐. Solubility increases.
AuCl 3 -‐ Cl-‐^ is the conjugate base of the strong acid HCl. Solubility is NOT affected. PbF 2 -‐ F-‐^ is the conjugate base of the weak acid HF. Solubility increases. ZnCO 3 -‐ CO 32 -‐^ is the conjugate base of the weak acid HCO 3 -‐. Solubility increases. 49. Would each of the following reagents increase, decrease, of have no effect on the solubility of Cu(OH) 2? CuCl 2 HCl NH 3 NaOH Cu(OH) 2 (s) Cu2+(aq) + 2OH-‐(aq) CuCl 2 -‐common ion, therefore decreases solubility. HCl-‐ H+^ neutralizes OH-‐, decreasing the concentration of OH-‐^ and shifting the equilibrium to the right, increasing solubility. NH 3 -‐ Cu2+^ forms the complex ion Cu(NH 3 ) 4 2+, increasing solubility. NaOH-‐ common ion, therefore decreases solubility. 50. Which salt solubility would be most sensitive to pH? Ca(NO 3 ) 2 CaF 2 CaCl 2 CaBr 2 CaI 2 All of the cations are Ca2+, the conjugate salt of the strong base Ca(OH) 2. It cannot be neutralized by changes in pH. Of all the anions, only F-‐^ is the conjugate base of a weak acid. The rest are conjugate bases of strong acids. Therefore CaF 2 is most sensitive to pH. 51. The solubility of which of the listed salts would be unaffected by the presence of a strong acid? KClO 4 BaF 2 FePO 4 SnI 2 Conjugate bases of strong acids will be unaffected by addition of a strong acid. The salts with this type of conjugate base include KClO 4 (HClO 4 is a strong acid) and SnI 2 (HI is a strong acid). BaF 2 and FePO 4 contain conjugate bases of weak acids and are therefore influenced by a strong acid addition. 52. How many of the following reagents, when added to a solution in contact with solid NiCO 3 would change the solubility of NiCO 3? HCl NiCl 2 NaCl Na 2 CO 3 NH 3 NiCO 3 (s) Ni2+(aq) + CO 32 -‐(aq) HCl-‐ This strong acid can neutralize CO 32 -‐^ because this anion is the conjugate base of a weak acid. This removes CO 32 -‐, shifting the reaction to the right and increasing solubility. NiCl 2 -‐ common ion, therefore decreases solubility. NaCl-‐ no effect. Na 2 CO 3 -‐ common ion, therefore decreases solubility. NH 3 -‐ Ni+^ forms the complex ion Ni(NH 3 ) 6 2+, increasing solubility.
57. For which salt should the aqueous solubility be most sensitive to pH? a) Ca(NO 3 ) 2 b) CaF 2 c) CaCl 2 d) CaBr 2 e) CaI 2 All of the cations are Ca2+, the conjugate salt of the strong base Ca(OH) 2. It cannot be neutralized by changes in pH. Of all the anions, only F-‐^ is the conjugate base of a weak acid. The rest are conjugate bases of strong acids. Therefore CaF 2 is most sensitive to pH.
Precipitation and Separation of Ions (Section 17.6)
To determine whether or not precipitation occurs, you need to compare Ksp to Q , the reaction quotient. The equation for Q appears the same as that for Ksp: Q= [products] [reactants] However, these concentrations can be at any point and do not necessarily correspond to equilibrium concentrations, as they do for Ksp. Ksp can be thought of as a specific Q value that occurs at equilibrium. Q for given concentrations can be compared to Ksp to determine whether or not precipitation will occur: **If Q<Ksp, there are not enough products. The solution is unsaturated , and the reaction will shift to the right, so NO precipitate is formed. More solid can be dissolved. **If Q=Ksp, the reaction is at equilibrium. The solution is saturated. **If Q>Ksp, there are too many products. The solution is supersaturated, and the reaction will shift to the left, so a precipitate will form. 58. A solution contains Ca2+^ at a concentration of 2.0 x 10-‐^4 M. If 40.0 mL of this solution is added to 25.0 mL of 5.0 x 10-‐^3 M NaF, will a precipitate form? If 40.0 mL of the Ca2+^ solution is added to 25.0 mL of 5.0 x 10-‐^3 M Na 3 PO 4 will a precipitate form? For CaF 2 , Ksp = 3.9 x 10-‐^11 For Ca 3 (PO 4 ) 2 , Ksp = 2.0 x 10-‐^29 CaF 2 (s) Ca2+(aq) + 2F-‐(aq) The concentrations of each ion must be calculated because the volumes change once they are added together: [Ca2+]= (2.0 x 10-‐^4 mol/L)(0.040 L)=1.23 x 10-‐^4 M (0.040 L + 0.0250 L) [F-‐]=( 5 .0 x 10-‐^3 mol/L)(0.0 25 L)=1.92 x 10-‐^3 M (0.040 L + 0.0250 L) Q=[Ca2+][F-‐]^2 =(1.23 x 10-‐^4 M)( 1.92 x 10-‐^3 M)^2 =4.55 x 10-‐^10 Q>Ksp, therefore equilibrium will shift left and a precipitate forms.
Ca 3 (PO 4 ) 2 (s) 3Ca2+(aq) + 2PO 43 -‐(aq) [Ca2+]= (2.0 x 10-‐^4 mol/L)(0.040 L)=1.23 x 10-‐^4 M (0.040 L + 0.0250 L) [PO 43 -‐]=( 5 .0 x 10-‐^3 mol/L)(0.0 25 L)=1.92 x 10-‐^3 M (0.040 L + 0.0250 L) Q=[Ca2+]^3 [ PO 43 -‐]^2 =(1.23 x 10-‐^4 M)^3 ( 1.92 x 10-‐^3 M)^2 =6.86 x 10-‐^18 Q>Ksp, therefore equilibrium will shift left and a **precipitate forms.
- Will a precipitate form when 4.5 mL of 0.025 M Pb(NO 3 ) 2 and 1.5 mL of 0.0065 M KF are mixed? How could you tell?** PbF 2 (s) Pb2+(aq) + 2F-‐(aq) [Pb2+]= (0.025 mol/L)(0.0045 L)=0.01875 M (0.0045 L + 0.0015 L) [F-‐]=(0.0065 mol/L)(0.0 015 L)=0.001625 M (0.0045 L + 0.0015 L) Q=[Ca2+][F-‐]^2 =(0.01875 M)(0.001625 M)^2 =4.95 x 10-‐^8 Q>Ksp, therefore equilibrium will shift left and a **precipitate forms.
- Which salt precipitates first and what is the minimum concentration of Ag+**^ necessary to cause this precipitation in a solution containing 7.5 mL of 0.025 M NaCl and 7.5 mL of 0.025 M Na 3 PO 4? Ksp AgCl = 1.8 x 10-‐^10 Ksp Ag 3 PO 4 = 1.3 x 10-‐^20 To determine which salt precipitates first, we need to look at Q, not Ksp, because precipitation occurs when Q>Ksp, which is not at equilibrium. We can, however, set Q equal to Ksp and solve for the minimum concentration of the ion required for precipitation to occur. Whichever concentration is lower, that solid will precipitate first : AgCl(s) Ag+(aq) + Cl-‐(aq) [Cl-‐]=(0.025 mol/L)(0.0075 L)=1.25 x 10-‐^2 M (0.0075 L + 0.0075 L) Q=[Ag+][Cl-‐] 1.8 x 10-‐^10 =[Ag+](1.25 x 10-‐^2 M) [Ag+]=1.4 x 10 -‐^8 M needed for AgCl to ppt. Ag 3 PO 4 (s) 3Ag+(aq) + PO 43 -‐(aq) [PO 43 -‐]=(0.025 mol/L)(0.0075 L)=1.25 x 10-‐^2 M (0.0075 L + 0.0075 L) Q=[Ag+]^3 [PO 43 -‐] 1.3 x 10-‐^20 =[Ag+]^3 (1.25 x 10-‐^2 M) [Ag+]=1.0 x 10 -‐^6 M needed for Ag 3 PO 4 to ppt. *Since the concentration of Ag+^ needed for AgCl to precipitate is lower, it will precipitate first.
64. A solution contains 0.005 M AsO 43 -‐, 0.005 M I-‐, and 0.005 M CO 32 -‐. If AgNO 3 is slowly added, in what order would the silver salts precipitate? For Ag 3 AsO 4 , Ksp = 1.0 x 10-‐^22 For AgI, Ksp = 8.3 x 10-‐^17 For Ag 2 CO 3 , Ksp = 8.1 x 10-‐^12 Ag 3 AsO 4 (s) 3Ag+(aq) + AsO 43 -‐(aq) Q=[Ag+]^3 [AsO 43 -‐] 1.0 x 10-‐^22 =[Ag+]^3 (0.005 M) [Ag+]=2.71 x 10 -‐^7 M needed for Ag 3 AsO 4 to ppt. AgI(s) Ag+(aq) + I-‐(aq) Q=[Ag+][I-‐] 8.3 x 10-‐^17 =[Ag+](0.005 M) [Ag+]=1.66 x 10 -‐^14 M needed for AgI to ppt. Ag 2 CO 3 (s) 2Ag+(aq) + CO 32 -‐(aq) Q=[Ag+]^2 [CO 32 -‐] 8.1 x 10-‐^12 =[Ag+]^2 (0.005 M) [Ag+]=4.02 x 10 -‐^5 M needed for Ag 2 CO 3 to ppt. ***Lowest concentration will precipitate first: AgI precipitates first, then Ag 3 AsO 4 , then Ag 2 CO 3.
- Three beakers contain the following solutions:
- 40.0 mL of 0.020 M Ca(NO 3 ) 2
- 40.0 mL of 0.020 M Fe(NO 3 ) 2
- 40.0 mL of 0.020 M Pb(NO 3 ) 2 If 10.0 mL of 0.050 M NaF is added to each beaker, in which beakers will a precipitate form? For CaF 2 , Ksp = 1.5 x 10-‐**^10 , FeF 2 Ksp = 2.4 x 10-‐^6 , PbF 2 Ksp = 7.1 x 10-‐^7 MF 2 (s) M2+(aq) + 2F-‐(aq) [M2+]= (0.020 mol/L)(0.040 L)=0.016 M (0.040 L + 0.010 L) [F-‐]=(0.050 mol/L)(0.010 L)=0.010 M (0.040 L + 0.010 L) Q=[M2+][F-‐]^2 =(0.016 M)(0.010 M)^2 =1.6 x 10-‐^6 Q=1.6 x 10-‐^6 for each equilibrium because the concentrations are the same. Compare the Q value to the individual Ksp values: CaF 2 : Q>Ksp, therefore equilibrium will shift left and a precipitate forms. FeF 2 : Q<Ksp, therefore equilibrium will shift right and no precipitate forms. PbF 2 : Q>Ksp, therefore equilibrium will shift left and a precipitate forms.
66. A solution contains 0.010 M Al3+^ and 0.010 M Ag+. Solid Na 3 PO 4 is slowly added to separate the two cations. Ksp for AlPO 4 is 1.3 x 10-‐^20 and Ksp for Ag 3 PO 4 is 1.3 x 10-‐^20. Which cation would precipitate first, and after it precipitates, what concentration of PO 43 -‐^ ion should be obtained in the solution for the best separation? AlPO 4 (s) Al3+(aq) + PO 43 -‐(aq) Q=[Al3+][PO 43 -‐] 1.3 x 10-‐^20 =(0.010 M)[PO 43 -‐] [PO 43 -‐]=1.3 x 10 -‐^18 M needed for AlPO 4 to ppt. Ag 3 PO 4 (s) 3Ag+(aq) + PO 43 -‐(aq) Q=[Ag+]^3 [PO 43 -‐] 1.3 x 10-‐^20 =(0.010 M)^3 [PO 43 -‐] [PO 43 -‐]=1.3 x 10 -‐^14 M needed for Ag 3 PO 4 to ppt. *Since the concentration of PO 43 -‐^ needed for AlPO 4 to precipitate is lower, it will precipitate first. To ensure the best separation, [PO 43 -‐] should be as close to the higher concentration (1.3 x 10-‐^14 M) WITHOUT going over that concentration. This ensures that virtually all of AlPO 4 will be solid, but no Ag+^ **will have precipitated yet into Ag 3 PO4.
- A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba2+. BaF 2 (Ksp = 1.7 x 10-‐**^6 ) will begin to precipitate when the concentration of F-‐^ ions reaches what value? Neglect volume changes associated with the addition of NaF solution. BaF 2 (s) Ba2+(aq) + 2F-‐(aq) Q=[Ba2+][F-‐]^2 1.7 x 10-‐^6 =(0.0144 M)[ F-‐]^2 [F-‐]=0.011 M needed for BaF 2 to ppt. 68. AgNO 3 is slowly added to a solution containing the following anions: 0.02 M AsO 42 -‐^ 0.02 M I-‐^ 0.02 M CO 32 -‐ In what order will these salts precipitate? Ksp for Ag 3 AsO 4 = 1.0 x 10-‐^22 AgI = 8.3 x 10-‐^17 Ag 2 CO 3 = 8.1 x 10-‐^12 Ag 3 AsO 4 (s) 3Ag+(aq) + AsO 43 -‐(aq) Q=[Ag+]^3 [AsO 43 -‐] 1.0 x 10-‐^22 =[Ag+]^3 (0.02 M) [Ag+]=1.7 x 10 -‐^7 M needed for Ag 3 AsO 4 to ppt. AgI(s) Ag+(aq) + I-‐(aq) Q=[Ag+][I-‐] 8.3 x 10-‐^17 =[Ag+](0.02 M) [Ag+]=4.2 x 10 -‐^15 M needed for AgI to ppt. Ag 2 CO 3 (s) 2Ag+(aq) + CO 32 -‐(aq) Q=[Ag+]^2 [CO 32 -‐] 8.1 x 10-‐^12 =[Ag+]^2 (0.02 M) [Ag+]=2.0 x 10 -‐^5 M needed for Ag 2 CO 3 to ppt. *Lowest concentration will precipitate first: AgI precipitates first, then Ag 3 AsO 4 , then Ag 2 CO 3.
