Question

The Molar Solubility of silver sulphate is 1.5×10^-2mol/L.The Solubility product (Ksp) of the salt will be::

Answer 1

Silver sulphate
Ag2(SO4)
$solubility \: product \: \\ = {2}^{2} \times {1}^{2} \times {s}^{3} = 4 {s}^{3} \\ where \: s \: is \: solubility \\ hence \: ksp = 4 {s}^{3} \\ 4 \times {(1.5 \times {10}^{ - 2}) }^{3} \\ = 13.5 \times {10}^{ - 6}$

Answer 2

Answer: The solubility product of the given salt is $1.35\times 10^{-5}$

Explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is represented as $K_{sp}$

Silver sulfate is an ionic compound formed by the combination of 2 silver ions and 1 sulfate ions.

The equilibrium reaction for the ionization of silver sulfate follows the equation:

$Ag_2SO_4(aq.)\rightarrow 2Ag^+(aq.)+SO_4^{2-}(aq.)$

The solubility product for the above reaction is:

$K_{sp}=[Ag^{+}]^2\times [SO_4^{2-}]$

We are given:

Solubility of silver sulfate = $1.5\times 10^{-2}mol/L$

So,

$[SO_4^{2-}]=1.5\times 10^{-2}mol/L$

$[Ag^+]=(2\times 1.5\times 10^{-2})=3.0\times 10^{-2}mol/L$

Putting values in above equation, we get:

$K_{sp}=(3.0\times 10^{-2})^2\times (1.5\times 10^{-2})\\\\K_{sp}=1.35\times 10^{-5}$

Hence, the solubility product of the given salt is $1.35\times 10^{-5}$