# chemistry

posted by john

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.

1. DrBob222

(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.

## Similar Questions

1. ### Chemistry!

0.0100 Ba(OH)2 is dissolved in water to give 500.0 mL of solution calculate OH- concentration would it be the molarity of Ba(OH)2 (0.0100 / 0.500) multiplied by two because there is twice as much OH?
2. ### chemistry

12 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.08 M C2H3O2-. What is the pH of the resulting solution?
3. ### chemistry

8 mL of 0.0100 M HCl are added to 22 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
4. ### chemistry

mL of 0.0100 M NaOH are added to 25.0 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
5. ### chemistry

Common ion effect;At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100?
6. ### chem101

If a solution contains 0.0100 mole of HCl in 10.0 mL aqueous solution. a.) What is its molarity?
7. ### science

10 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.1 M C2H3O2-. What is the pH of the resulting solution?
8. ### Science (chemistry) help please

The following data was obtained from titrating a solution containing 25 mL of 0.0100 M MgCl2 solution, 1 mL of NH3OH buffer, and 10 drops of Calmagite with EDTA. Calculate the concentration of EDTA. Report the answer with three sig …
9. ### chemistry

10 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.1 M C2H3O2-. What is the pH of the resulting solution?
10. ### Chemistry

12 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.11 M C2H3O2-. What is the pH of the resulting solution?

More Similar Questions