we prepare a solution by mixing 0.10L of 0.12M sodium chloride with 0.23 L of a 0.18M MgCl2 solution. what volume of a 0.20 M silver nitrate solution do we need to precipitate all the cl- ion int the solution as AgCl.
V = 0.10 L
M = 0.12 NaCl
V2 = 0.23 L
M = 0.18 MgCl2
V3 = ? AgNO3
recipitate all Cl- ions
We must go to equilibrium
Ag+(aq) + Cl-(aq) <---> AgCl(s)
Ksp = [Ag+][Cl-]
[Cl-] = ions of NaCl and ions of MgCl2
find each amount
moles of NaCl = M*V = 0.1*0.12 = 0.012 mol of NaCl present
moles of MgCl2 = M*V = 0.23*0.18 = 0.0414 mol of MgCl2 present (we can see that Cl is present double)
Total moles of Cl = 0.012+ 2*0.0414 = 0.0948 mol of Cl-
Ag+(aq) + Cl- <---> AgCl(s)
since Ksp is so small (1-8*10^-10) we can say that there is almost no solubility
therefore, we need 0.0948 mol of Ag+ to react with 0.0948 mol of Cl-
therefore
AgNO3 ---> Ag+ andNO3-
we need 0.0948 mol of AgNO3
Find that in volume of solution
M = mol/V
V = mol/M = 0.0948 mol / 0.20 = 0.474 liters are needed
or 474 ml
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