Question

67. Solubility of AgCl 0.1 M CaCl2 solution will be
(Ksp of AgCl = 1.8 * 10-10)
(1) 1.8 * 10-10
(2) 9 10-10
(3) 1.8 x 10-11
(4) 9 * 10-11​

Answer 1

WE address the equilibrium...

A

g

+

+

C

l

−

⇌

A

g

C

l

(

s

)

Explanation:

For which

K

sp

=

[

A

g

+

]

[

C

l

−

]

=

?

?

We are given that

[

A

g

C

l

]

=

1.9

×

10

−

4

⋅

g

143.32

⋅

g

⋅

m

o

l

−

1

0.100

⋅

L

=

1.33

×

10

−

5

⋅

m

o

l

⋅

L

−

1

...

But by the stoichiometry...

[

A

g

+

]

=

[

C

l

−

]

=

1.33

×

10

−

5

⋅

m

o

l

⋅

L

−

1

And thus

K

sp

=

[

1.33

×

10

−

5

]

2

=

1.76

×

10

−

10

...the which is close enuff to your answer....

Happy?

Answer 2

Option (1) is the correct answer

Solubility of AgCl 0.1 M CaCl2 solution will be Ksp= 1.8×$10^{10}$

Explanation:

Solubility of AgCl 0.1 M CaCl2 solution will be determined.

Here AgCl will dissociate as

AgCl -----Ag+  + Cl-

(s)            (s)        (s) +0.1

CaCl ----- Ca+   + 2 Cl-

0.1                       0.1+s

Ksp[AgCl]= [Ag+] {Cl-]

Given Ksp of AgCl = 1.8×$10^{10}$

1.8×$10^{10}$ =[0.1][s+0.1]

s= 1.8×$10^{10}$ /0.1×10

s= 1.8×$10^{10}$

• In Solubility the bond formation takes place between the solute molecules and solvent molecules.
• When quantity is taken care of, The maximum concentration of solute that dissolves is the known concentration of solvent at a given temperature.
• The solubility product constant is denoted by Ksp​,
• It is the equilibrium constant of a solid substance which is dissolved in an aqueous solution.
• Solubility product represents the amount at which a solute dissolves in solution.
• If the substance is more soluble, the higher will be the Ksp value.

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