Question

If the solubility product of a salt A2B3 is 1.08 x 10^-23.
then its solubility in the aqueous solution will be
a) 2.7 x 10^-12 mol L^-1
b) 54 x 10^-5 mol L-1
c) 1 x 10^-8 mol L^-1
d) 1 x 10^-5 mol L^-1​

Answer 1

Answer:

c) 1 x 10^-8 mol L^-1 . answer ho jaye

Answer 2

Answer:

The solubility of the salt $A_{2}B_{3}$ in the aqueous solution is $10^{-5}mol/L$.

Therefore, option d) $1\times 10^{-5}mol/L$ is correct.

Explanation:

Given,

The solubility product of the salt $A_{2}B_{3}$, $K_{sp}$ = $1.08\times 10^{-23}$

The solubility in the aqueous solution =?

Let the solubility of $A_{2}B_{3}$ in aqueous solution = S

Now, the dissociation of salt into ions in an aqueous solution:

•          $A_{2}B_{3} \rightarrow 2A^{3+} + 3B^{2-}$

 Solubility:  S          2S       3S

Now, from the reaction, the solubility product:

• $K_{sp}$ = $[A^{3+}]^{2} [B^{2-}]^{3}$
• $1.08\times 10^{-23}$ = $[2S]^{2}[3S]^{3}$
• $1.08\times 10^{-23}$ = $108S^{5}$
• $S^{5}$ = $10^{-25}$
• S = $10^{-5}$ mol/L

Hence, the solubility of salt in the aqueous solution = $10^{-5}mol/L$.