10) Common ion effect question. The Na2CO3 affects initial conc of carbonate
Ksp = 8.46 x 10^-12
Ag2CO3(s) <--> 2Ag+(aq) + CO3 2-(aq)
Ksp = [Ag+]^2 [CO3 2-]
| Ag+ | CO32- |
| R | 2 | 1 |
| I | 0 | 0.125 |
| C | +2x | +x |
| E | 2x | 0.125 + x |
8.46 x 10^-12 = (2x)^2 (0.125+x)
As K is small, small x approximation can be made, so 0.125 + x = 0.125
8.46 x 10^-12 = 4x^2 (0.125)
1.692 x 10^-11 = x^2
x = 4.11 x 10^-6 mol/L
(there may be a mistake in the answers provided, A doesn't take into the 2 molar ratio of Ag+)
23) This Q type of question doesn't really work well with a RICE table because you have to factor in dilutions that occur when two solutions are mixed. We are finding if a precipitate is made in the final solution and so you can't just use the conc in 150 mL and conc in 100 mL.. You need to find the conc of each in the diluted 250 mL solution
Possible precipitate here is PbCl2
PbCl2(s) <--> Pb2+(aq) + 2Cl-(aq)
Q = [Pb2+][Cl-]^2
The molar ratios here are not really important, it's just about finding the Pb2+ conc from Pb(NO3)2 and Cl- conc from NaCl and substituting.
For Pb2+:
c1v1 = c2v2
(0.100)(150) = c2(250)
c2 = 0.06 mol/L ==> [Pb2+] = 0.06 mol/L (as only 1 Pb2+ in Pb(NO3)2)
For NaCl:
c1v1 = c2v2
(0.200)(100) = c2(250)
c2 = 0.08 mol/L ==> [Cl-] = 0.08 mol/L (as only 1 Cl- in NaCl)
Q = (0.06)(0.08)^2
Q = 3.84 x 10^-4
Ksp = 1.70 x 10^-5
Q>Ksp, so equilibrium shifts left and precipitate will be made
