1 L of 0.002 mol/L Pb(NO3)2 has been prepared. Given that ksp for PbCl2 is 1.7 x10^-5, what is the largest number of moles of nacl that can be added to this solution without causing a precipitate to form.
Ksp PbCl2 = 1.7E-5 = (Pb^2+)(Cl^-)^2
You know Ksp and (Pb^2+), substitute and solve for (Cl^-). That gives the molarity of Cl^- which = M NaCl and since that is in 1 L solution that is mols NaCl.
I have a similar question and I got 1.7x10^-8 but would we have to multiply this value by 2?
To determine the largest number of moles of NaCl that can be added to the solution without causing a precipitate to form, we need to compare the solubility products (Ksp) of the two compounds: Pb(NO3)2 and PbCl2.
Given:
- Concentration of Pb(NO3)2: 0.002 mol/L
- Ksp for PbCl2: 1.7 x 10^-5
First, let's write the balanced equation for the dissociation of PbCl2:
PbCl2 (s) ⇌ Pb2+ (aq) + 2Cl- (aq)
The Ksp expression for PbCl2 is:
Ksp = [Pb2+][Cl-]^2
Since the stoichiometric coefficients in the equation are 1:2:1, and the concentration of Pb(NO3)2 is 0.002 mol/L, the concentration of Pb2+ is also 0.002 mol/L.
Using the Ksp expression, we can calculate the concentration of Cl- ions at equilibrium:
1.7 x 10^-5 = (0.002)(x)^2
Solving the equation for x:
x^2 = (1.7 x 10^-5) / (0.002)
x^2 = 8.5 x 10^-3
x = √(8.5 x 10^-3)
x ≈ 0.0922 mol/L
This concentration represents the maximum concentration of Cl- ions that can be present in the solution without causing a precipitate to form. Remember that NaCl dissociates completely in water, giving one Na+ and one Cl- ion for each formula unit.
Therefore, the maximum number of moles of NaCl that can be added is equal to the concentration of Cl- ions:
Maximum moles of NaCl = 0.0922 mol/L
Please note that this value assumes ideal behavior and neglects potential common ion effects.