OWL question so its 3 parts Part A A solution contains 9.25×10-3 M aluminum nitr
ID: 994286 • Letter: O
Question
OWL question so its 3 parts
Part A
A solution contains 9.25×10-3 M aluminum nitrate and 5.90×10-3 M copper(II) acetate. Solid sodium hydroxide is added slowly to this mixture. What is the concentration of aluminum ion when copper(II) ion begins to precipitate?
[Al3+] = M
Part B
A solution contains 1.29×10-2 M iron(II) acetate and 9.60×10-3 M manganese(II) nitrate.
Solid potassium hydroxide is added slowly to this mixture.
What is the concentration of iron(II) ion when manganese(II) ion begins to precipitate?
[Fe2+] = M
Part C
A solution contains 1.02×10-2 M ammonium carbonate and 5.02×10-3 M sodium phosphate.
Solid silver nitrate is added slowly to this mixture.
What is the concentration of phosphate ion when carbonate ion begins to precipitate?
[phosphate] = M
Explanation / Answer
A) The Ksp values are:
Ksp , Al(OH)3 : 1.9*10-33
Ksp , Cu(OH)2 : 1.6*10-19
Let us calculate the concentration of OH- when Cu(II) begins to precipitate. The dissociation of Cu(OH)2 is given as
Cu(OH)2 (s) <=====> Cu2+ (aq) + 2 OH- (aq)
Q = [Cu2+][OH-]2 (concentration of a solid is neglected)
When precipitation occurs, Q = Ksp.
Therefore,
Ksp = [Cu2+][OH-]2
or, 1.6*10-19 = (5.90*10-3)[OH-]2
or, [OH-]2 = 2.71*10-17 ===> [OH-] = 5.21*10-9
Hence, [OH-] when Cu(II) begins to precipitate is 5.21*10-9 M.
Now, we use this concentration in the expression for Ksp for Al(OH)3.
Al(OH)3 (s) <======> Al3+ (aq) + 3 OH- (aq)
Ksp = [Al3+][OH-]3
==> 1.9*10-33 = [Al3+](5.21*10-9)3
or. 1.9*10-33 = [Al3+](1.41*10-25)
or, [Al3+] = 1.9*10-33/1.41*10-25 = 1.35*10-8
The concentration of Al3+ is 1.35*10-8 M (ans)
B) Ksp , Fe(OH)2 = 4.9*10-17
Ksp , Mn(OH)2 = 2.1*10-13
The dissociation of Mn(OH)2 is given as
Mn(OH)2 (s) <=====> Mn2+ (aq) + 2 OH- (aq)
Ksp = [Mn2+][OH-]2
When Mn(II) begins to preceipitate,
Ksp = (9.60*10-3)[OH-]2
or, 2.1*10-13 = (9.60*10-3)[OH-]2
==> [OH-]2 = 2.19*10-11
or, [OH-] = 4.68*10-6
The concentration of OH- is 4.68*10-6.
To calculate the concentration of Fe(II) at this time, we write the dissociation equation and note that
Fe(OH)2 (s) <====> Fe2+ (aq) + 2 OH- (aq)
Ksp = [Fe2+][OH-]2 ==> 4.9*10-17 = [Fe2+](4.68*10-6)2
or, 4.9*10-17 = [Fe2+](2.19*10-11) ==> [Fe2+] = 2.24*10-6
The concentration of Fe(II) is 2.24*10-6 M (ans)
C) Ksp , Ag2CO3 : 8.1*10-12
Ksp , Ag3PO4 : 8.9*10-17
The dissociation of silver carbonate occurs as
Ag2CO3 (s) <====> 2 Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
When carbonate begins to precipitate, we have
8.1*10-12 = [Ag+]2(1.02*10-2) ==> [Ag+] = 2.82*10-5
The silver ion concentration is 2.82*10-5 M
The dissociation of Ag3PO4 follows
Ag3PO4 (s) <=====> 3 Ag+ (aq) + PO43- (aq)
Ksp = [Ag+]3[PO43-]
or. 8.9*10-17 = (2.82*10-5)3[PO43-]
==> [PO43-] = 3.97*10-3
The concentration of phosphate ion is 3.97*10-3 M (ans)