Calculate the K S P (Solubility product), if solubility of A g C N = 7.73 × 10 − 9 M

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Answer 1

Question

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

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Answer 2

Question

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

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Answer 3

Question

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

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Answer 4

Question

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

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Answer 5

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

Answer 6

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

Answer 7

Chapter 12, Problem 12.64PAE

(a)

Interpretation Introduction

To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

Answer 8

(a)

Expert Solution

Explanation of Solution

AgCN(S)⇌Ag+(aq)+CN−(aq)KSP=[Ag+][CN−]

Solubility of AgCN is 7.73×10−9M, it means that

Concentration of Ag+([Ag+]) at equilibrium =1×7.73×10−9M.

Concentration of CN−([CN−]) at equilibrium =1×7.73×10−9M.

[Ag+]=7.73×10−9M[CN−]=7.73×10−9MKSP=[Ag+][CN−]=7.73×10−9×7.73×10−9=5.98×10−17

Hence KSP of CNCN=5.98×10−17.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

Answer 13

(b)

Interpretation Introduction

To determine: Calculate KSP of Ni(OH)2. If solubility is 5.98×10−17M.

Answer 14

(b)

Expert Solution

Explanation of Solution

Ni(OH)2(S)⇌Ni2++2OH−

1 Mole of Ni(OH)2 gives 1 mole Ni+0 and 2 moles OH−.

Solubility Ni(OH)2 is 5.16×10−6M.

[Ni2+]=1× solubility of Ni(OH)2=5.16×10−6M[OH−]=2× Solubility of Ni(OH)2=2×(5.16× 10 −6M)=10.32×10−6M

From the definition of solubility product.

KSP=[Ni2+][OH−]2

Putting the value in R.H.S. you get

=(5.16× 10 −6M)(10.32× 10 −6M)=5.50×10−16M

Hence solubility product of Ni(OH)2=5.50×10−16M.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

Answer 19

(c)

Interpretation Introduction

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

Answer 20

(c)

Expert Solution

Explanation of Solution

Cu3(PO4)2(S)⇌3Cu2+(aq)+2PO42−(aq).

From the above balanced equilibrium reaction it becomes clear that 1 mole of Cu3(PO4)2(S) gives 3

Cu2+(aq) and 2 moles of PO42−(aq).

The solubility of Cu3(PO4)2(S) is 1.67×10−8M.so at equilibrium

[Cu2+]=3× slubility of Cu3(P O 4)2[PO42−]=2× solubility of Cu3(P O 4)2

So,

[Cu2+]=3×1.67×10−8M[PO42−]=2×1.67×10−8M

From the definition of solubility product.

KSP=[Cu 2+]3[PO4 2−]2=(5.01× 10 −8)3(3.34× 10 −8)23=1.40×10−37

Hence solubility product of Cu3(PO4)2 is 1.40×10−37.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Conclusion

You have learnt that how to find the KSP from the solubility of a given salt.

Answer 25

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Calculate the K S P (Solubility product), if solubility of A g C N = 7.73 × 10 − 9 M

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To determine: Calculate the KSP (Solubility product), if solubility of AgCN=7.73×10−9M

Explanation of Solution

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Explanation of Solution

To determine: Calculate KSP of Cu3(PO4)2. Solubility is 1.67×10−8M.

Explanation of Solution

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