Taking the Ksp of Zn(OH)2 as 4.0 x 10-3 and the Kf of [Zn(NH3)6] 2+ as 9.9 x 108 , determine how many moles of Zn(OH)2 can dissolve in 0.6 L of 1.2 M NH3.
Taking the Ksp of Zn(OH)2 as 4.0 x 10-3 and the Kf of [Zn(NH3)6] 2+ as 9.9 x 108 , determine how many moles of Zn(OH)2 can dissolve in 0.6 L of 1.2 M NH3.
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Taking the Ksp of Zn(OH)2 as 4.0 x 10-3 and the Kf of [Zn(NH3)6] 2+ as 9.9 x 108 , determine how many moles of Zn(OH)2 can dissolve in 0.6 L of 1.2 M NH3.
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![2h. Taking the Ksp of Zn(OH)2 as 4.0 x 10-³ and the Kƒ of [Zn(NH3)6] 2+ as 9.9 x 108
determine how many moles of Zn(OH)2 can dissolve in 0.6 L of 1.2 M NH3.
equilibrium reaction that is associated with the kap for In (OH) ₂
Zn(OH)₂ (s) = Znay (aq) + 20H⁰² (aq)
Ksp = [2n²+ ] [04²] 2
equilibrium reaction that is associated with the kf for [Zn(NH₂)]2+
[2n (NH₂)]2+ (aq)
2
24
2n (99)
[2n (NH₂)]²+
(Zn) [N3]
▶ When add these two reaction.
Kf
I
C
=
Zn(OH)₂ (s) + Zn²+ (aq) + 6NM₂ (aq) = 3₁²+ (aq) + 201² (99) + [2n (NH3)]2+ (94)
Zn(OH)₂ (5) + 6NH ₂ (aq) = [2n(NH₂)6]2+ (99) + 201² (99)
1.2
O
+2x
2x
+ 6 NH3 raa)
E
Kc = Ksp x Kf
6
= 4×10 ¹³ x 9.9×108
= 3.96×106
Hence kc >> |
3.96 X106
3.96x10
62 <<<1
3.96x106
-6x
1-2-6x
=
= (x) (2x)²²
(1-2-62)6
= 4
[Zn(NH₂) 6 ]²+ [OH-] 2
[NH₂] 6
=
(1.2-676)²)
-→ 1.2-6x ≈ 1.2
423
1.26
3
x=143.5183911
+x
x
molar solubility of Zn(OH)₂
= 143.52 M
mole aissolved
143 52 x 0.6
86.11 mols
=](https://content.bartleby.com/qna-images/question/c0fdaa3d-4c74-4e89-abd4-5bdd8a4f6bb4/324bfcdf-5ac8-44d0-9ce6-3bc642c89d76/gug905o_processed.jpeg)
Transcribed Image Text:2h. Taking the Ksp of Zn(OH)2 as 4.0 x 10-³ and the Kƒ of [Zn(NH3)6] 2+ as 9.9 x 108
determine how many moles of Zn(OH)2 can dissolve in 0.6 L of 1.2 M NH3.
equilibrium reaction that is associated with the kap for In (OH) ₂
Zn(OH)₂ (s) = Znay (aq) + 20H⁰² (aq)
Ksp = [2n²+ ] [04²] 2
equilibrium reaction that is associated with the kf for [Zn(NH₂)]2+
[2n (NH₂)]2+ (aq)
2
24
2n (99)
[2n (NH₂)]²+
(Zn) [N3]
▶ When add these two reaction.
Kf
I
C
=
Zn(OH)₂ (s) + Zn²+ (aq) + 6NM₂ (aq) = 3₁²+ (aq) + 201² (99) + [2n (NH3)]2+ (94)
Zn(OH)₂ (5) + 6NH ₂ (aq) = [2n(NH₂)6]2+ (99) + 201² (99)
1.2
O
+2x
2x
+ 6 NH3 raa)
E
Kc = Ksp x Kf
6
= 4×10 ¹³ x 9.9×108
= 3.96×106
Hence kc >> |
3.96 X106
3.96x10
62 <<<1
3.96x106
-6x
1-2-6x
=
= (x) (2x)²²
(1-2-62)6
= 4
[Zn(NH₂) 6 ]²+ [OH-] 2
[NH₂] 6
=
(1.2-676)²)
-→ 1.2-6x ≈ 1.2
423
1.26
3
x=143.5183911
+x
x
molar solubility of Zn(OH)₂
= 143.52 M
mole aissolved
143 52 x 0.6
86.11 mols
=
Solution
by Bartleby ExpertFollow-up Question
how does x/0.6 cpme from?
and x is moles or concentration in step 4?
Solution
by Bartleby ExpertKnowledge Booster
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