Question

For a sparingly soluble salt AB2, if solubility is x M , then it's solubility product will be?

Answer 1

Aloha!

Solubility process is:

1

AB

2

→

S

A

2+

+

2S

2B

−

Using the relation of solubility and solubility product as:

K

sp

=[A

2+

][B

−

]

2

K

sp

=S.(2S)

2

Substituting S=10

−5

M

K

sp

=10

−5

×(2×10

−5

)

2

K

sp

=4×10

−15

Therefore option A is correct.

Answer 2

Answer:

The solubility of the product is 4x³.

Explanation:

The ionization of the salt is given as,

$AB_2\rightarrow A+2B$          (1)

And the solubility product will be given as,

$K_s_p=[A][B]^2$       (2)

Where,

$K_s_p$=solubility product

[A]=concentration of A

[B]=concentration of B

The solubility of the salt is "x" M.

So,

[A]=x

[B]=2x

By substituting all the required values in equation (2) we get;

$K_s_p=[x][2x]^2$

$K_s_p=4x^3$

Hence, the solubility of the product is 4x³.

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