Solubility Homework Set Worked Out Solutions

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

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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...
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