Solubility Homework Set Worked Out Solutions

352m 37x102m xm xm 4xm 37x102xm xm 23524xm 2nh

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: ial Change Equilibrium CdCO3(s) Cd2+(aq) + CO32‐(aq) Ksp=5.2 x 10‐12 Cd(NH3)42+(aq) Kf=1.3 x 107 Cd2+(aq) + 4NH3(aq) 2+(aq)+ CO 2‐(aq) K= K x K CdCO3(s) + 4NH3 (aq) Cd(NH3)4 3 sp f ‐12 7 =(5.2 x 10 )(1.3 x 10 ) ‐5 =6.76 x 10 CdCO3(s) + 4NH3(aq) Cd(NH3)42+(aq) + CO32‐(aq) ‐‐‐‐‐‐ 1.5 M 0 M 0 M ‐‐‐‐‐‐ ‐4x M +x M +x M ‐‐‐‐‐‐ 1.5‐4x M x M x M K=[Cd(NH3)42+][CO32‐]= ___x2____ = 6.76 x 10‐5 [NH3]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 NH3? Kf Co(NH3)62+ = 8.3 x 104 Co2+(aq) + 6NH3(aq) Co(NH3)62+(aq) Initial 4.75 x 10‐2 M 5.0 M 0 M Change ‐4.75 x 10‐2 M ‐6(4.75 x 10‐2) M +4.75 x 10‐2 M Final (NOT equilibrium) 0 M 4.715 M 4.75 x 10‐2 M Co(NH3)62+(aq) Co2+(aq) + 6NH3(aq) 4.75 x 10‐2 M 0 M 4.715 M ‐x M +x M +6x M 4.75 x 10‐2 ‐x M xM 4.715+6x M 2+][NH ]6= (x)(4.715+6x)6 = _(1.1 x 104)(x)_=___1_____ Kd=[Co 3 [Co(NH3)62+] (4.75 x 10‐2 ‐x) 4.75 x 10‐2 8.3 x 104 x= 5.2 x 10‐11 M Initial Change Equilibrium 37. Calculate the molar solubility of AgCl in 12 M NH3. Kf Ag(NH3)2 = 1.7 x 107 Ksp AgCl = 1.8 x 10‐10 Initial Change Equilibrium AgCl(s) Ag+(aq) + Cl‐(aq) Ksp=1.8 x 10‐10 Ag(NH3)2+(aq) Kf=1.7 x 107 Ag+(aq) + 2NH3(aq) AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl‐(aq) K= Ksp x Kf =3.06 x 10‐3 AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl‐(aq) ‐‐‐‐‐‐ 12 M 0 M 0 M ‐‐‐‐‐ ‐2x M +x M +x M ‐‐‐‐‐‐ 12‐2x M x M x M K=[Ag(NH3)2+][Cl‐]= ___x2____ = 3.06 x 10‐3 [NH3]2 (12‐2x)2 x= 6.6 x 10‐1 M 10 Dr. Fus CHEM 123 38. Calculate the concentration of free aluminum ion, [Al3+], in 1.0 L of solution that contains 0.040 mol Al(NO3)3 and 2.00 mol NaF. Kf for AlF63‐ is 7.1 x 1019. Al3+(aq) + 6F‐(aq) Initial 0.040 M 2.00 M Change ‐0.040 M ‐6(0.040) M Final (NOT equilibrium) 0 M 1.76 M AlF63‐(aq) 0 M +0.040 M 0.040 M AlF63‐ (aq) Al3+(aq) + 6F‐(aq) 0.040 M 0 M 1.76 M ‐x M +x M +6x M 0.040‐x M xM 1.76+6x M Kd=[Al3+][F‐]6= (x)(1.76+6x)6 = _(2.97 x 101)(x)_=___1_____ [AlF63‐] (0.040 ‐x) 0.040 7.1 x 1019 x= 1.9 x 10‐23 M Initial Change Equilibrium 39. The Cd2+ ion forms the complex ion CdCl42‐ for which Kf = 6.3 x 102. Determine the equilibrium constant for the solubility of CdCO3 in contact with a solution that contains Cl‐ ion. Ksp for CdCO3 is 5.2 x 10‐12. CdCO3(s) Cd2+(aq) + CO32‐(aq) 2+(aq) + 4Cl‐(aq) CdCl42‐ (aq) Cd CdCO3(s) + 4Cl‐(aq) CdCl42‐(aq)+ CO32‐(aq) Ksp=5.2 x 10‐12 Kf=6.3 x 102 K= Ksp x Kf ‐12 2 =(5.2 x 10 )(6.3 x 10 ) ‐9 K=3.28 x 10 40. Use information from problem 39 to calculate the molar solubility of CdCO3 in 3.0 M NaCl. Initial Change Equilibrium CdCO3(s) + 4Cl‐(aq) ‐‐‐‐‐‐ 3.0 M ‐‐‐‐‐‐ ‐4x M ‐‐‐‐‐‐ 3.0‐4x M CdCl42‐(aq) + CO32‐(aq) 0 M 0 M +x M +x M x M x M K=[CdCl42+][CO32‐]= ____x2____ = 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...
View Full Document

Ask a homework question - tutors are online