In: Chemistry
1.The molar solubility of iron(III) sulfide in a water solution is_________M
2.The equilibrium concentration of hydroxide ion in a saturated silver hydroxide solution is__________M
3.The equilibrium concentration of silver ion in a saturated silver chromate solution is___________M
4.A student measures the molar solubility of lead
phosphate in a water solution to be
7.86×10-10 M.
Based on her data, the solubility product constant for this
compound is________
5.A student measures the molar solubility of barium
phosphate in a water solution to be
6.48×10-7 M.
Based on her data, the solubility product constant for this
compound is________
6.The solubility of Mg(OH)2 is measured and found to be 9.29×10-3 g/L. Use this information to calculate a Ksp value for magnesium hydroxide. Ksp=________?
7.The solubility of Ag2S is measured and found to be 7.12×10-15 g/L. Use this information to calculate a Ksp value for silver sulfide. Ksp=________?
8.The solubility of ZnCO3 is measured and found to be 5.04×10-4 g/L. Use this information to calculate a Ksp value for zinc carbonate. Ksp=________?
1)
Fe2S3 ----> 2 Fe+3 + 3 S-2
the solubility product constant is given by
Ksp = [Fe+3]^2 [S-2]^3
now
Fe2S3 ---> 2 Fe+3 + 3 S-2
let the molar solubility be s
then
[Fe+3] = 2s
[S-2] = 3s
now
Ksp = [Fe+3]^2 [S-2]^3
Ksp = [2s]^2 [3s]^3
Ksp = 108 s^5
we know that
Ksp value of Fe2S3 is 1.4 x 10-88
so
1.4 x 10-88 = 108 x s^5
s = 1.0533 x 10-18
so
the molar solubility of iron (III) sulfide is 1.0533 x 10-18 mol / L
2)
AgOH ----> Ag+ + OH-
Ksp = [Ag+] [OH-] = 2 x 10-8
now
AgOH ---> Ag+ + OH-
let the molar solubility of AgOH be s
then at equilibrium
[Ag+] = s
[OH-] = s
now
Ksp = [Ag+] [OH-]
Ksp = [s] [s]
Ksp = s2
s2 = 2 x 10-8
s = 1.414 x 10-4
so
the equilibrium concentration of Hydroxide ion is 1.414 x 10-4 M
3)
Ag2Cr04 ----> 2 Ag+ + Cr042-
Ksp = [Ag+]^2 [Cr042-] = 1.12 x 10-12
now
Ag2Cr04 ----> 2 Ag+ + Cr042-
let the molar solubility of Ag2Cr04 be s
then at equilibrium
[Ag+] = 2s
[Cr042-] = s
so
Ksp = [Ag+]^2 [Cr042-]
Ksp = [2s]^2 [s]
Ksp = 4s3
4s3 = 1.12 x 10-12
s =6.542 x 10-5
now
[Ag+] = 2s
[Ag+] = 2 x 6.542 x 10-5
[Ag+] = 1.3084 x 10-4
so
the equilibrium concentration of Ag+ ion is 1.3084 x 10-4 M
4)
Pb3(PO4)2 ---> 3 Pb+2 + 2 P043-
let the molar solubility be s
then
[Pb+2] = 3s
[P043-] = 2s
now
Ksp = [Pb+2]^3 [P043-]^2
Ksp = [3s]^3 [2s]^2
Ksp = 108s^5
given
molar solubility (s) = 7.86 x 10-10
so
Ksp = 108 x (7.86 x 10-10)^5
Ksp = 3.118 x 10-44
so
the value of solubility product constant is 3.118 x 10-44
5)
Ba3(PO4)2 ---> 3 Ba+2 + 2 P043-
let the molar solubility be s
then
[Ba+2] = 3s
[P043-] = 2s
now
Ksp = [Ba+2]^3 [P043-]^2
Ksp = [3s]^3 [2s]^2
Ksp = 108s^5
given
molar solubility (s) = 6.48 x 10-7
so
Ksp = 108 x (6.48 x 10-7)^5
Ksp = 1.234 x 10-29
so
the value of solubility product constant is 1.234 x 10-29
6)
we know that
molar solubility = solubility in g/L / molar mass
so
molar solubility of Mg(OH)2 = 9.29 x 10-3 / 58.3197
molar solubility of Mg(OH)2 = 1.593 x 10-4 M
now
Mg(OH)2 ----> Mg+2 + 2 OH-
let the molar solubility of Mg(OH)2 be s
then
[Mg+2] = s
[OH-] = 2s
now
Ksp = [Mg+2] [OH-]^2
Ksp = [s] [2s]^2
Ksp = 4s3
Ksp = 4 x ( 1.593 x 10-4)^3
Ksp = 1.617 x 10-11
so
the Ksp value of magnesium hydroxide is 1.617 x 10-11
7)
we know that
molar solubility = solubility in g/L / molar mass
so
molar solubility of Ag2S = 7.12 x 10-15 / 247.8
molar solubility of Ag2S = 2.8733 x 10-17 M
now
Ag2S ----> 2 Ag+ + S-2
let the molar solubility of Ag2S be s
then
[Ag+] = 2s
[S-2] = s
now
Ksp = [Ag+]^2 [S-2]
Ksp = [2s]^2 [s]
Ksp = 4s3
Ksp = 4 x ( 2.8733 x 10-17)^3
Ksp = 9.49 x 10-50
so
the Ksp value of silver sulfide is 9.49 x 10-50
8)
molar solubility = solubility in g/L / molar
mass
so
molar solubility of ZnC03 = 5.04 x 10-4 / 125.39
molar solubility of ZnC03 - 4.02 x 10-6 M
now
ZnC03 ---> Zn+2 + C032-
let the molar solubility be s
then
[Zn+2] = s
[C032-] = s
now
Ksp = [Zn+2] [C032-]
Ksp = [s] [s]
Ksp = s2
Ksp = ( 4.02 x 10-6)^2
Ksp = 1.6156 x 10-11
so
the value of Ksp for zinc carbonate is 1.6156 x 10-11