1) Solid ammonium sulfide is slowly added to 75.0mL of a 0.0542M calcium acetate
ID: 959027 • Letter: 1
Question
1) Solid ammonium sulfide is slowly added to 75.0mL of a 0.0542M calcium acetate solution. The concentration of sulfide ion required to just initiate precipitation is _____ M.
2) Solid magnesium acetate is slowly added to 50.0mL of a 0.0463M sodium fluoride solution. The concentration of magnesium ion required to just initiate precipitation is _____ M. Solid ammonium sulfide is slowly added to 75.0 ml of a 0.0542 M calcium acetate solution. The concentration of sulfide ion required to just initiate precipitation is M.
Explanation / Answer
Precipitation occurs when Ionic product(Qp) > Solubility product(KsP)
It means the minimum concentration of ions for which Ksp will come out will be the just start of precipitation.
1.For CaS we want to find out [S] = ?
For CaS we write,
CaS < ------------ > Ca2+ + S2-
Let S be the solubilty of CaS in moles/L and hence [Ca2+] = S mole/L and [S2-] = S moles/L.
Ksp is defined as,
Ksp = [Ca2+][S2-]
Ksp =S x S
Ksp = S2.
Ksp = 8 x 10-6.
S2 = 8x 10-6
S = 2.83 x 10-3 moles/L
i.e. [S2-] = S = 2.83 x 10-3 moles/L
This is the concentration of Sulphide ions to just initiate the precipitation. Above this concentration precipitation will certainly occur.
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2) for, MgF2 Magnesium fluoride, Ksp = 6.4 x 10-9
For MgF2 we write,
MgF2 -----------> Mg2+ + 2 F-
Let S be the solubility of MgF2 salt hence, [Mg2+]= S moles/L and [F-] = 2S moles/L
Ksp = [Mg2+][F-]2
Ksp = (S) x (2S)2
Ksp = 4 S3
4S3 = Ksp
4S3 =6.4 x 10-9
S3 = (6.4 x 10-9) / 4
S3 = 1.6 x 10-9
S = (1.6 x 10-9)1/3
S = 1.17 x10-3 moles/L
Hence [F-] = 2 x S = 2 x 1.17 x 10-3 moles/L
[F-]= 2.34 x 10-3 moles/L
This is the minimum fluoride ion concentration to precipitation to just start above this concentration precipitation will certainly occur.
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