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Question: The volume of water needed to dissolve \(1g\) of \(BaS{O_4}\) \(({K_{sp}} = 1.1 \times {10^{ - 10}})...

The volume of water needed to dissolve 1g1g of BaSO4BaS{O_4} (Ksp=1.1×10−10)({K_{sp}} = 1.1 \times {10^{ - 10}}) at 250C{25^0}C is:
A.100lit100lit
B.200lit200lit
C.410lit410lit
D.600lit600lit

Explanation

Solution

At first think about the definition of solubility product of the solution. The solubility product constant is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in a solution.

Complete step by step answer:
Ksp{K_{sp}} is the solubility product constant. The solubility product constant for barium sulphate is,
Ksp=[Ba+2][SO4−2]{K_{sp}} = [B{a^{ + 2}}][S{O_4}^{ - 2}]
The dissociation of barium sulphate solution is,
BaSO4⇌Ba2++SO42−BaS{O_4} \rightleftharpoons B{a^{2 + }} + S{O_4}^{2 - }
Let’s take the concentration of barium ion and sulphate ion as xx and xx respectively.
Ksp=[x][x]{K_{sp}} = [x][x]

1.1×10−10=x2 x=1.05×10−5moles/litre  1.1 \times {10^{ - 10}} = {x^2} \\\ x = 1.05 \times {10^{ - 5}}moles/litre \\\
In 1litre1litre, 1.05×10−5moles1.05 \times {10^{ - 5}}moles
In xlitresxlitres,1233moles\dfrac{1}{{233}}moles
Where molar mass of the barium sulphate is 233g233g
Then by calculating xx, we get
x=11.05×10−5×233 x=408.7litres  x = \dfrac{1}{{1.05 \times {{10}^{ - 5}} \times 233}} \\\ x = 408.7litres \\\
Hence the required volume of water is 408.7litres408.7litres. So the correct answer is C.

Additional Information:- The solubility equilibrium is an example of dynamic equilibrium in that some individual molecules migrate between the solid and solution phases such that the rates of dissolution and precipitation are equal to one another. When equilibrium is established, the solution is said to be saturated. The concentration of the solute in the saturated solution is known as solubility. Solubility is temperature dependent. A solution containing a higher concentration of solute than the solubility is said to be supersaturated. Solubility will increase with decreasing size of solute particles because of the additional surface energy.

Note:
The more soluble a substance is, the value of the Ksp{K_{sp}} will be higher . A solubility equilibrium exists when a chemical compound in the solid state is in chemical equilibrium with a solution containing the compound.