Solubility Homework Set Worked Out Solutions

Solubilityincreases

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online