Chemistry help (1 Viewer)

[aspired]

Can anyone explain to me how to do this question?

Equal volumes of lead nitrate solution and sodium iodide solution, each with a concentration
of 0.10mol L-1, are mixed. A yellow precipitate of lead iodide results.
Pb2+ + 2I - > PbI2 (s)
Calculate the concentration of each ion remaining in solution.

 

[Eightyseven87]

^ Got the same question, apparently according to our teachers most of the schools had the same test.

I also would like to know the answer since I just skipped it lol

 

[iSplicer]

EDIT Please type up the entire question next time in your first post... just giving the solid prec. wont help as the concentration of each will be ZERO. I need complete eq

 

[aspired]

Hrmm.. this was the solution
I don't understand it

 

[iSplicer]

Hrmm.. this was the solution
I don't understand it

You didn't provide me with the whole question in the first post... =( you missed all traces of Na

 

[aspired]

Sorry, that was the whole question my friend gave me ><

 

[richman92]

I think the solution has something wrong
here my solution :
I convert everything to mole
Can you guys just check it

Because the Volumes of Pb(NO3)2 and NaI are equal, let them be 10(L)
then

* Before reaction :
moles of Pb(NO3)2 = 10*0.1 = 1(mole) => moles of Pb(2+) is 1 mole and NO3(-) is 2 moles
moles of NaI = 10*0.1 = 1(mole) => moles of Na(+) and I(-) are equal to 1 mole

* Reaction
Pb(2+) + 2I(-) --> PbI2(s)
Initial 1 mole 1mole 0 mole
Change o.5 mole 1 mole 0.5 mole ( Pb(2+) is excess )
Final 0.5 mole 0 0.5 mole

[Pb(2+)] = n/V = o.5/(10+10) = 0.025M
[I(-)] = 0

Because Na(+) and NO3(-) do not form precipitate then their moles remain the same as before reaction

[Na(+)] = n/V = 1/(10+10) = 0.05M
[NO3(-)] = n/v = 2/(10+10) = 0.1M

 

[Kaatie]

this is very similar to working out equilibrium constants in industrial chem in year 12

 

[richman92]

plz check my solution
I thik it's correct as I can not find any mistake