Question

the solubility product of A2B is 32 × 10 ^ - 9 mol^3 L^-1. calculate its solubility.please answer this

Answer 1

The solubility of $A_2B$ is $2\times 10^{-3}moles/liter$

Explanation:

The equation for the reaction will be as follows:

$A_2B\rightleftharpoons 2A^++B^-$

By Stoichiometry,

1 mole of  $A_2B$ gives 2 moles of $A$ and 1 mole of B

Thus if solubility of $A_2B$  is s moles/liter, solubility of A is 2s moles\liter and solubility of B is s moles/liter

Therefore,  

$K_sp=[2A^{+}]^2[B^{-}]$

$K_sp=[2s]^2[s]$

$4s^3=32\times 10^{-9}mol^3L^{-1}$

$s=2\times 10^{-3}moles/liter$

Hence, the solubility of $A_2B$ is $2\times 10^{-3}moles/liter$

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