Asked by himani.
the fictitious compound pandemonium carbonate has a ksg=3.091*10^-9 M 3 in water at poor temperature .calculate the solubility of PnCo3 in an aqueous solution of 1.11 M pandemonium sulfate.
express your answer in units of molarity.
express your answer in units of molarity.
Answers
Answered by
DrBob222
.......PnCO3 ==> Pn^2+ + CO3^2-
I......solid.....1.11.....0
C....x dissolves..x.......x
E......solid......x.......x
Ksp = (Pn^2+)(CO3^2-)
3.091E-9 = (x+1.11)(x)
Solve for x = solubility of PnCO3 in mols/L.
I......solid.....1.11.....0
C....x dissolves..x.......x
E......solid......x.......x
Ksp = (Pn^2+)(CO3^2-)
3.091E-9 = (x+1.11)(x)
Solve for x = solubility of PnCO3 in mols/L.
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