What you've done is almost correct assuming the spacing is ok. The numbers are but some are in the wrong place. This forum doesn't do spacing very well. I'll try to improve that.
.......... AgCl ==> Ag^+ + Cl^-
I.........solid.....0.......0
C.........solid.....x.......x
E.........solid.....x.......x
Then the AgNO3 ionizes completely as follows:
............AgNO3 ==> Ag^+ + NO3^-
E....................6.5E-3
Ksp = 1.8E-10 = (Ag^+)(Cl^-)
(Ag^) from AgCl is x
(Ag^+) from AgNO3 is 6.5E-3
Total is 6.5E-3 + x so the equation looks this way.
1.8E-10 = (6.5E-3+x)(x)
You have solved this the easy way by assuming 6.5E-3+x = 6.5E-3 and I think that probably is ok. You could go through the quadratic and see. At any rate this is the solubility of AgCl in mols/L. Convert that to grams/L by
grams AgCl = mols AgCl x molar mass AgCl.
Calculate solubility of AgCl (in g/L) in a 6.5*10^-3 M silver Nitrate solution ksp= 1.8 * 10^-10
(AgCl molecular mass is 143.3g) any idea on how to do this this is what i tried>
AgCl--> Ag + Cl
I 6.5*10^-3 0
C same x
E same x
ksp = [Ag][Cl]
(1.8 * 10^-10) = (6.5 * 10^-3)x
x = 2.77 * 10^-8 M
I don't know what to do next please help me thank you
1 answer
1 answer