Information Set 4:
Consider the reaction between 500 mL of 0.100 mol L–1 Pb(NO3)2(aq) and 500 mL of 0.200 mol L–1Na2CrO4(aq).
Ksp for PbCrO4 (s) is 2.8 x 10–13.
What volume of 0.100 mol L–1 lead nitrate solution would give the necessary concentration of lead ion to start the formation of a precipitate?
I think the question is worded wrong because 500 mL of Pb(NO3)2 and 500 mL Na2CrO4 will ppt PbCrO4 big time. I think the problem is asking for volume of 0.1M Pb(NO3)2 needed to start pptn of PbCrO4 in 500 mL of 0.2M Na2CrO4.l
If you start with 0.2M Na2CrO4, then (CrO4^2-) = 0.2 M so
(Pb^2+)(CrO4^2-) = Ksp
Solve for (Pb^2+) = about 1E-12M.
Then (0.1M Pb^2+ x ?L = approx 1E-12 so ?L is a very small number. Technically one would need to add the volumes in because the volume added of Pb(NO3)2 would dilute that 0.2M Na2CrO4; however, since the volume of Pb ion needed is so small we can forget about the added volume acting as a diluent.
To find the volume of the 0.100 mol L–1 lead nitrate solution that would give the necessary concentration of lead ions to start the formation of a precipitate, we need to use the concept of the common ion effect and the solubility product constant (Ksp).
The balanced chemical equation for the reaction between lead nitrate (Pb(NO3)2) and sodium chromate (Na2CrO4) is:
Pb(NO3)2 + Na2CrO4 → PbCrO4(s) + 2NaNO3
From the equation, we can see that the stoichiometric ratio between Pb(NO3)2 and PbCrO4 is 1:1. This means that for every 1 mole of Pb(NO3)2, we will get 1 mole of PbCrO4.
Given that the Ksp for PbCrO4 is 2.8 x 10–13, we can write the solubility product expression as:
Ksp = [Pb2+][CrO42-]
Since the stoichiometry of the reaction is 1:1, the concentration of Pb2+ ions in the solution will be the same as the concentration of PbCrO4, and we can substitute [Pb2+] with the concentration of PbCrO4 in the solubility product expression:
Ksp = [PbCrO4]^2
Solving for [PbCrO4], we get:
[PbCrO4] = √(Ksp)
[PbCrO4] = √(2.8 x 10–13)
[PbCrO4] ≈ 1.67 x 10–7 mol L–1
Therefore, to start the formation of a precipitate, the concentration of PbCrO4 must be at least 1.67 x 10–7 mol L–1.
Now, we can use the volume and concentration information provided for the sodium chromate solution (0.200 mol L–1, 500 mL) to calculate the amount of Pb(NO3)2 needed to reach the required concentration.
Concentration of Pb(NO3)2 = 0.100 mol L–1
Let's assume the volume of the lead nitrate solution needed is V mL.
The amount of Pb(NO3)2 used in the reaction is:
Amount of Pb(NO3)2 = (Concentration of Pb(NO3)2) × (Volume of Pb(NO3)2)
Amount of Pb(NO3)2 = (0.100 mol L–1) × (V mL / 1000) L
Since the stoichiometric ratio between Pb(NO3)2 and PbCrO4 is 1:1, the amount of Pb(NO3)2 should be equal to the amount of PbCrO4:
Amount of Pb(NO3)2 = Amount of PbCrO4
(0.100 mol L–1) × (V mL / 1000) L = (1.67 x 10–7 mol L–1) × (500 mL / 1000) L
Simplifying and solving for V, we get:
V = (1.67 x 10–7 mol L–1) × (500 mL / 1000) L / (0.100 mol L–1)
V ≈ 8.35 x 10–3 L
Finally, converting the volume to milliliters, we get:
V ≈ 8.35 mL
Therefore, approximately 8.35 mL of the 0.100 mol L–1 lead nitrate solution would give the necessary concentration of lead ion to start the formation of a precipitate.