Asked by melissa
The Ksp for silver sulfate (Ag2SO4) is 1.2 10-5. Calculate the solubility of silver sulfate in each of the following.
A)water
B)0.14 M AgNO3
C)0.33 M K2SO4
the answer in mol/L
A)water
B)0.14 M AgNO3
C)0.33 M K2SO4
the answer in mol/L
Answers
Answered by
DrBob222
This is the common ion effect. In part a there is no common ion, part b as the Ag+ as the common ion, and part c has the SO4^= as the common ion.
Ag2SO4 ==> 2Ag^+ + SO4^=
..x........2x........x
Ksp = (Ag^+)^2(SO4^=)
Substitute Ksp, Ag^+ = 2x and SOr^= = x. Solve for x = solubility
b) AgNO3 ==> Ag^+ + NO3^-
....0.14M.....0......0
equil..0......0.14...0.14
Use Ksp as before but not
(Ag^+) 0.14 + 2x
(SO4^=) = x
Solve for x
c) K2SO4 ==> 2K^+ + SO4^=
...0.33........0.....0
.....0........0.66...0.33
Use Ksp as before but
(Ag^+) = 2x
(SO4^=) = x + 0.33
Solve for x
Post your work if you get stuck.
For A,
Ag2SO4 ==> 2Ag^+ + SO4^=
..x........2x........x
Ksp = (Ag^+)^2(SO4^=)
Substitute Ksp, Ag^+ = 2x and SOr^= = x. Solve for x = solubility
b) AgNO3 ==> Ag^+ + NO3^-
....0.14M.....0......0
equil..0......0.14...0.14
Use Ksp as before but not
(Ag^+) 0.14 + 2x
(SO4^=) = x
Solve for x
c) K2SO4 ==> 2K^+ + SO4^=
...0.33........0.....0
.....0........0.66...0.33
Use Ksp as before but
(Ag^+) = 2x
(SO4^=) = x + 0.33
Solve for x
Post your work if you get stuck.
For A,
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