To a 100.0 mL solution containing 0.0100 M Ba(NO3)2 and 0.0100 M Pb(NO3)2 was added 101.0 mL of 0.0100 M H2SO4 to provide a slight excess of SO42- relative to Ba2+ in the resulting solution. Assuming that:
i) H+ does not bind to SO42-
ii) ionic strength effects can be ignored
iii) the system immediately achieves equilibrium
and given that, under the above conditions:
i) Ksp (BaSO4) = 1.00 × 10-10 M2
ii) Ksp (PbSO4) = 1.70 × 10-8 M2
Determine the following:
a) The concentrations of Pb2+, Ba2+ and SO42- in the final solution.
b) The composition of the precipitate, reported as the mole fraction of BaSO4 and the mole fraction of PbSO4.
(Ba^2+)(SO4^2-) = 1E-10
(Pb^2+)(SO4^2-) = 1.7E-8
How many mmols Ba(NO3)2 do you have. That's 1 mmol.
How much Pb(NO3)2 = 1 mmol.
(Ba^2+) initially = 1 mmol/301 mL = 3.22E-3M
(Pb^2+) = 3.22E-3M
If you start dripping H2SO4 drop wise into the solution the BaSO4 will start pptng first. What is the (SO4^2-) at that point? That's 1E-10/3.33E-3 = 3.1E-8. BaSO4 will continue to ppt as each drop H2SO4is added until Ksp for PbSO4r is reached.What must the (SO4^2-) be when Pb^2+ starts to ppt?
(SO4^2-) = 1.7E-8/3.22E-3 = 5.28E-6 M.
What is the (Ba^2+) at this point? It is (Ba^2+) = 1E-10/5.28E-6 = 1.89E-5 M. Convert that to millimols Ba^2+ in solution, subtract from 1 mmol initially to find BaSO4 pptd and convert to mols BaSO4.
Calculate mmols H2SO4 used to ppt the BaSO4 to that point and subtract from 1.01 mmols H2SO4 initially to see how much H2SO4 is left. Then write
Pb^2+ + SO4^2- ==> PbSO4
Knowing you have 1 mmols Pb and 0.016 mmols SO4, I would assume all of the SO4 is used to form PbSO4. Convert that to mols PbSO4 and calculate mole fraction PbSO4 and XBaSO4 from that.
To fine Pb^2+ at the end, 1 mmols Pb - ? mmols SO4 = mmols Pb^2+ left.
All of that gives you Ba^2+, Pb^2+, SO4^2- and XBaSO4 and XPbSO4. Post your work if you get stuck.
To determine the concentrations of Pb2+, Ba2+, and SO42- in the final solution, we need to consider the stoichiometry of the reaction between Ba(NO3)2, Pb(NO3)2, and H2SO4. Let's go step by step to solve this problem.
Step 1: Calculate the moles of Ba(NO3)2, Pb(NO3)2, and H2SO4
Given:
- Volume of Ba(NO3)2 and Pb(NO3)2 solution = 100.0 mL = 0.100 L
- Concentration of Ba(NO3)2 and Pb(NO3)2 = 0.0100 M
- Volume of H2SO4 solution added = 101.0 mL = 0.101 L
- Concentration of H2SO4 = 0.0100 M
Moles of Ba(NO3)2 = Concentration x Volume = 0.0100 M x 0.100 L = 0.0010 moles
Moles of Pb(NO3)2 = Concentration x Volume = 0.0100 M x 0.100 L = 0.0010 moles
Moles of H2SO4 = Concentration x Volume = 0.0100 M x 0.101 L = 0.0010 moles
Step 2: Write the balanced equation for the reaction between Ba(NO3)2, Pb(NO3)2, and H2SO4
Ba(NO3)2 + Pb(NO3)2 + H2SO4 → BaSO4 + PbSO4 + 2 HNO3
Step 3: Determine the limiting reagent
From the balanced equation, we can see that the stoichiometric ratio between Ba(NO3)2, Pb(NO3)2, and H2SO4 is 1:1:1. Since the moles of each reactant are the same (0.0010 moles), none of them is limiting.
Step 4: Calculate the moles of products formed
Moles of BaSO4 formed = Moles of Ba(NO3)2 = 0.0010 moles
Moles of PbSO4 formed = Moles of Pb(NO3)2 = 0.0010 moles
Step 5: Calculate the concentrations of Pb2+, Ba2+, and SO42- in the final solution
For Pb2+:
- Volume of final solution = Volume of initial solution + Volume of H2SO4 added = 0.100 L + 0.101 L = 0.201 L
- Concentration of Pb2+ in the final solution = Moles of Pb2+ / Volume of final solution
= 0.0010 moles / 0.201 L = 0.004975 M
For Ba2+:
- Concentration of Ba2+ in the final solution = Moles of Ba2+ / Volume of final solution
= 0.0010 moles / 0.201 L = 0.004975 M
For SO42-:
- The slight excess of SO42- is provided by the added H2SO4, which is in excess compared to Ba2+.
- Therefore, the concentration of SO42- in the final solution is the same as the initial Ba(NO3)2 solution, which is 0.0100 M.
So, in the final solution:
- Concentration of Pb2+ = 0.004975 M
- Concentration of Ba2+ = 0.004975 M
- Concentration of SO42- = 0.0100 M
Now, let's move on to part b) and determine the composition of the precipitate:
Moles of BaSO4 formed = 0.0010 moles
Moles of PbSO4 formed = 0.0010 moles
Mole fraction of BaSO4 = Moles of BaSO4 formed / Total moles of precipitate
= 0.0010 moles / (0.0010 moles + 0.0010 moles) = 0.5
Mole fraction of PbSO4 = Moles of PbSO4 formed / Total moles of precipitate
= 0.0010 moles / (0.0010 moles + 0.0010 moles) = 0.5
Therefore, the composition of the precipitate is 50% BaSO4 and 50% PbSO4, reported as mole fractions.
Note: The assumption made in this problem (iii) that the system immediately achieves equilibrium might not hold true in a practical scenario. However, if the conditions described are met, these calculations provide an approximation.