
The volume of water needed to dissolve $1g$ of $BaS{O_4}$$({K_{sp}} = 1.1 \times {10^{ - 10}})$ at ${25^0}C$ is:
A.$100lit$
B.$200lit$
C.$410lit$
D.$600lit$
Answer
510.6k+ views
Hint: 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:
${K_{sp}}$ is the solubility product constant. The solubility product constant for barium sulphate is,
${K_{sp}} = [B{a^{ + 2}}][S{O_4}^{ - 2}]$
The dissociation of barium sulphate solution is,
$BaS{O_4} \rightleftharpoons B{a^{2 + }} + S{O_4}^{2 - }$
Let’s take the concentration of barium ion and sulphate ion as $x$ and $x$ respectively.
${K_{sp}} = [x][x]$
$
1.1 \times {10^{ - 10}} = {x^2} \\
x = 1.05 \times {10^{ - 5}}moles/litre \\
$
In $1litre$, $1.05 \times {10^{ - 5}}moles$
In $xlitres$,$\dfrac{1}{{233}}moles$
Where molar mass of the barium sulphate is $233g$
Then by calculating $x$, we get
$
x = \dfrac{1}{{1.05 \times {{10}^{ - 5}} \times 233}} \\
x = 408.7litres \\
$
Hence the required volume of water is $408.7litres$. 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 ${K_{sp}}$ 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.
Complete step by step answer:
${K_{sp}}$ is the solubility product constant. The solubility product constant for barium sulphate is,
${K_{sp}} = [B{a^{ + 2}}][S{O_4}^{ - 2}]$
The dissociation of barium sulphate solution is,
$BaS{O_4} \rightleftharpoons B{a^{2 + }} + S{O_4}^{2 - }$
Let’s take the concentration of barium ion and sulphate ion as $x$ and $x$ respectively.
${K_{sp}} = [x][x]$
$
1.1 \times {10^{ - 10}} = {x^2} \\
x = 1.05 \times {10^{ - 5}}moles/litre \\
$
In $1litre$, $1.05 \times {10^{ - 5}}moles$
In $xlitres$,$\dfrac{1}{{233}}moles$
Where molar mass of the barium sulphate is $233g$
Then by calculating $x$, we get
$
x = \dfrac{1}{{1.05 \times {{10}^{ - 5}} \times 233}} \\
x = 408.7litres \\
$
Hence the required volume of water is $408.7litres$. 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 ${K_{sp}}$ 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.
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