Question

How many grams of $Ca{C_2}{O_4}$ will dissolve in distilled water to make one liter of a saturated solution. (${K_{sp}} = 2.5 \times {10^{ - 9}}$ and its molecular mass is $128$)
A) $0.0064g$
B) $0.0128g$
C) $0.0032g$
D) $0.0640g$

Answer 1

Hint: Hint- In the given question the value of solubility product i.e. ${K_{sp}}$ is given as $Ca{C_2}{O_4}$. To find out the mass dissolved in the distilled water one needs to find out the number of moles based on given data. The value of molar mass is given which can be used to calculate the mass of the solute which is to be dissolved in distilled water.

Complete step by step answer:
1) First of all the value ${K_{sp}}$ is given from which we can find out the number of moles present in the solute which is being dissolved.
2) Now, let's see the solubility product constant for $Ca{C_2}{O_4}$ in which when the chemical compound $Ca{C_2}{O_4}$ will dissolve in distilled water to make one liter of saturated solution it forms cations and anions in the water. The reaction representation is given as follows,
$Ca{C_2}{O_4}\overset {} \leftrightarrows C{a^{2 + }} + {C_2}{O_4}^{2 - }$
3) Now let's assume the values of both anion and cation formed as $x$ then the ${K_{sp}}$ value becomes
${K_{sp}} = {\left[ x \right]^2}$
So the value of $x$ will be,
$x = \sqrt {{K_{sp}}} = \sqrt {2.5 \times {{10}^{ - 9}}} = 5 \times {10^{ - 5}}$
The value $x$ means the $5 \times {10^{ - 5}}$ number of moles $Ca{C_2}{O_4}$ will dissolve in distilled water to make one liter of a saturated solution.
4) As we now know the value of the number of moles we can now find out the value of mass which needed to be dissolved by the following formula,
${\text{Mass = number of moles }} \times {\text{ molar mass = 5}} \times {\text{1}}{{\text{0}}^{ - 5}} \times 128 = 0.0064g$
5) Therefore, the mass $0.0064g$ of $Ca{C_2}{O_4}$ will dissolve in distilled water to make one liter of saturated solution

which shows the option A as a correct choice.

Note:
The value given as ${K_{sp}}$ that is known as a solubility product which is a type of equilibrium constant and its value depends on temperature. The solubility product value usually increases with an increase in temperature due to increased solubility of the solute in the solvent. A saturated solution is a solution in which there will be no more dissolution of a solute in the solvent.