If .100 M NaCl (aq) is added slowly to a beaker containing both .120 M AgNO3 (aq) and .150 M Pb(NO3)2 (aq) at 25C which will precipitate first, AgCl(s)PbCl2(s)?
AgCl ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-)
(Cl^-) = Ksp/(Ag^+).
YOu know Ksp and Ag^+, solve for Cl^-
PbCl2 ==> Pb^2+ + 2Cl^-
Ksp = (Pb^2+)(Cl^-)^2
(Cl^-) = sqrt(Ksp/(Pb^2+)
You know Ksp and (Pb^2+), solve for Cl^-.
Now consider that you are dripping the NaCl into the beaker (slowly) containing BOTH salts, the first ppt will be the one that exceeds the Ksp first.
To determine which compound will precipitate first (AgCl or PbCl2), we need to compare their solubility products. The solubility product, also known as the equilibrium constant of solubility, is a measure of how soluble a compound is in water.
First, let's write the balanced chemical equations for the precipitation reactions:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
Next, let's write the solubility product expressions for AgCl and PbCl2:
For AgCl:
Ksp(AgCl) = [Ag+][Cl-]
For PbCl2:
Ksp(PbCl2) = [Pb2+][Cl-]^2
Now, let's calculate the concentrations of Ag+ and Pb2+ when the solutions are mixed. To do so, we need to consider the stoichiometry of the reactions and assume complete dissociation.
For AgCl:
- AgNO3 dissociates completely, so [Ag+] = 0.120 M.
- NaCl dissociates completely, so [Cl-] = [Na+] = 0.100 M.
For PbCl2:
- Pb(NO3)2 dissociates completely, so [Pb2+] = 0.150 M.
- NaCl dissociates completely, so [Cl-] = [Na+] = 0.100 M.
Now, let's substitute the concentrations into the solubility product expressions to find the solubility product values.
For AgCl:
Ksp(AgCl) = [Ag+][Cl-] = (0.120 M)(0.100 M) = 0.012
For PbCl2:
Ksp(PbCl2) = [Pb2+][Cl-]^2 = (0.150 M)(0.100 M)^2 = 0.0015
Comparing the solubility product values, we can see that Ksp(AgCl) > Ksp(PbCl2). Therefore, AgCl has a higher solubility product and is less soluble than PbCl2.
In this case, when .100 M NaCl is slowly added to the beaker containing .120 M AgNO3 and .150 M Pb(NO3)2, AgCl will precipitate first.