Write the hydrolysis reactions and calculate the solubility for the solid in each of the following solutions: (Use the appendix in your book for Ksp values.) For extra practice, figure out the formulas from the names…. 1. Iron(II) carbonate in water FeCO3(s) < == > Fe2+ + CO32‐ Ksp = [Fe2+][CO32‐] = 3.1 x 10‐11 Fe2+ CO32‐ I 0 0 C +s +s E s s 3.1 x 10‐11 = s2 5.6 x 10‐6 mol/L = s 2. Iron(II) carbonate in 0.030 M aqueous potassium carbonate K2CO3 → 2 K+ + CO32‐ FeCO3(s) < == > Fe2+ + CO32‐ Ksp = [Fe2+][CO32‐] = 3.1 x 10‐11 Fe2+ CO32‐ I 0 0.030 C +s +s E s 0.030+s 3.1 x 10‐11 = (s)(0.030 + s) assume s is small ‐11
3.1 x 10 = 0.030s 1.0 x 10‐9 mol/s = s 3. Calcium fluoride in water CaF2 < == > Ca2+ + 2 F‐ Ksp = [Ca2+][F‐]2 = 3.9 x 10‐11 Ca2+ F‐ I 0 0 C +s +2s E s 2s 3.9 x 10‐11 = (s)(2s)2 = 4s3 2.1 x 10‐4 mol/L = s 4. Calcium fluoride in 0.12 M aqueous calcium chloride CaCl2 → Ca2+ + 2 Cl‐ CaF2 < == > Ca2+ + 2 F‐ Ksp = [Ca2+][F‐]2 = 3.9 x 10‐11 Ca2+ F‐ I 0.12 0 C +s +2s E 0.12 + s 2s 3.9 x 10‐11 = (0.12 +s)(s)2 assume s is small ‐11
2
3.9 x 10 = 0.12s 1.8 x 10‐5 mol/L = s 5. Silver sulfide in water Ag2S < == > 2 Ag+ + S2‐ Ksp = [Ag+]2[S2‐] = 1.6 x 10‐49 S2‐ Ag+ I 0 0 C +2s +s E 2s s 1.6 x 10‐49 = (2s)2(s) = 4s3 3.4 x 10‐17 mol/L = s 6. Lead(II) iodide in 0.081 M aqueous calcium iodide CaI2 → Ca2+ + 2 I‐ PbI2 < == > Pb2+ + 2 I‐ Ksp = [Pb2+][I‐]2 = 8.5 x 10‐9 Pb2+ I‐ I 0 0.16 C +s +2s E s 0.16 + 2s 8.5 x 10‐9 = (s)(0.16 + 2s)2 assume 2s is small ‐9
2
8.5 x 10 = s(0.16) = 0.0262s 3.2 x 10‐7 mol/L = s 7. Lead(II) iodide in 0.023 M aqueous lead(II) nitrate Pb(NO3)2 → Pb2+ + 2 NO3‐ PbI2 < == > Pb2+ + 2 I‐ Ksp = [Pb2+][I‐]2 = 8.5 x 10‐9 Pb2+ I‐ I 0.023 0 C +s +2s E 0.023+s 2s 8.5 x 10‐9 = (0.023+s)(2s)2 assume s is small ‐9
2
2
8.5 x 10 = 0.023(4s ) = 0.092s 3.0 x 10‐4 mol/L = s 8. copper(II) phosphate in water Cu3(PO4)2 < == > 3 Cu2+ + 2 PO43‐ Ksp = [Cu2+]3[PO43‐]2 = 1.4 x 10‐37 Cu2+ PO43‐ I 0 0 C +3s +2s E 3s 2s 1.4 x 10‐37 = (3s)3(2s)2 = (27s3)(4s2) = 108s5 1.7 x 10‐8 mol/L = s