Solubility Homework Set Worked Out Solutions

# Solubilityincreases

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Unformatted text preview: ion AgCl2‐ for which Kf = 2.5 x 105. 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) AgCl2‐(aq) Ag+(aq) + 2Cl‐(aq) ‐(aq) AgCl(s) + Cl AgCl2‐ Ksp=1.8 x 10‐10 Kf=2.5 x 105 K= Ksp x Kf ‐5 K=4.5 x 10 42. Use the information from problem 41 to calculate the molar solubility of AgCl in 8.5 M HCl. Initial Change Equilibrium AgCl(s) + Cl‐(aq) AgCl2‐(aq) ‐‐‐‐‐‐ 8.5 M 0 M ‐‐‐‐‐‐ ‐x M +x M ‐‐‐‐‐‐ 8.5‐x M x M 11 Dr. Fus CHEM 123 K=[AgCl2‐]= ___x____ = 4.5 x 10‐5 [Cl‐] (8.5‐x) x= 3.8 x 10‐4 M 43. Calculate the concentration of free copper ion, [Cu2+], in a 1.0 L solution containing 3.7 x 10‐2 mol Cu2+ and 2.5 M NH3. Kf for Cu(NH3)42+ = 5 x 1012. Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq) 2.5 M 0 M Initial 3.7 x 10‐2 M Change ‐3.7 x 10‐2 M ‐4(3.7 x 10‐2) M +3.7 x 10‐2 M Final (NOT equilibrium) 0 M 2.352 M 3.7 x 10‐2 M Cu(NH3)42+(aq) Cu2+(aq) + 4NH3(aq) 0 M 2.352 M 3.7 x 10‐2 M ‐x M +x M +4x M 3.7 x 10‐2 ‐x M xM 2.352+4x M 2+][NH ]4= (x)(2.352+4x)4 = _(3.1 x 101)(x)_=___1_____ Kd=[Cu 3 [Cu(NH3)42+] (3.7 x 10‐2 ‐x) 3.7 x 10‐2 5.0 x 1012 x= 2.4 x 10‐16 M Initial Change Equilibrium 44. What is the concentration of Cd2+ in a solution that is 0.10 M Cd(NH3)42+ ? Kf for Cd(NH3)42+ is 4.0 x 106 Initial Change Equilibrium Cd(NH3)42+(aq) 0.10 M ‐x M 0.10‐x M Cd2+(aq) + 4NH3(aq) 0 M 0 M +x M +4x M xM 4x M =[Cd2+][NH 4= (x)(4x)4 = _256x5_=___1_____ Kd 3 [Cd(NH3)42+] (0.10‐x) 0.10 4.0 x 106 x= 1.0 x 10‐2 M 45. Calculate the molar solubility of ZnS in 2.5 M NaOH. Ksp ZnS = 3.0 x 10‐23 Kf Zn(OH)42‐ = 2.9 x 1015 ZnS(s) Zn2+(aq) + S2‐(aq) 2+(aq) + 4OH‐(aq) Zn(OH)42‐ (aq) Zn ZnS(s) + 4OH‐(aq) Zn(OH)42‐(aq)+ S2‐(aq) Ksp=3.0 x 10‐23 Kf=2.9 x 1015 K= Ksp x Kf K=8.7 x 10‐8 Initial Change Equilibrium ZnS(s) + 4OH‐(aq) ‐‐‐‐‐‐ 2.5 M ‐‐‐‐‐‐ ‐4x M ‐‐‐‐‐‐ 2.5‐4x M Zn(OH)42‐(aq) + S2‐(aq) 0 M 0 M +x M +x M x M x M K=[Zn(OH)42+][S2‐]= ____x2____ = 8.7 x 10‐8 [OH‐]4 (2,5‐4x)4 x= 1.8 x 10‐3 M 12 Dr. Fus CHEM 123 46. The formation constant, Kf, for Ni(NH3)62+ 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 NH3 (concentrations refer to the moment before the formation of the complex ion)? Ni2+(aq) + 6NH3(aq) Ni(NH3)62+(aq) Initial 0.045 M 3.0 M 0 M Change ‐0.045 M ‐6(0.045) M +0.045 M Final (NOT equilibrium) 0 M 2.73 M 0.045 M Ni(NH3)62+(aq) Ni2+(aq) + 6NH3(aq) 0.045 M 0 M 2.73 M ‐x M +x M +6x M 0.045 ‐x M xM 2.73 +6x M 2+][NH ]6= (x)( 2.73 +6x)6 = _(4.1 x 102)(x)_=___1_____ Kd=[Ni 3 [Ni(NH3)62+] (0.045 ‐x) 0.045 5.5 x 108 x= 2.0 x 10‐13 M Initial Change Equilibrium 47. A solution is saturated with silver acetate, AgC2H3O2 (Ksp = 1.9 x 10–3). Which of the following reagents will increase the solubility of silver acetate? HNO3 NH3 AgNO3 NaC2H3O2 For problems like these, first write out the solubility equilibriu...
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