The conductivity of 0.001M `Na_2SO_4` solution is `2.6xx10^(-4)Scm^(-1)` and increases to `7.0xx10^(-4)Scm^(-1)`, When the solution is saturated with `CaSO_4`. The molar conductivities of `Na^+` and `Ca^(2+)` are 50 and 120`Scm^2mol^(-1)`, respectively. Neglect conductivity of used water. What is the solubility product of `CaSO_4`?

Options

A. `4xx10^(-6)`
B. `1.57xx10^(-3)`
C. `4xx10^(-4)`
D. `1.52xx10^(-4) `

Answer

To solve the problem, we need to find the solubility product (Ksp) of CaSO4 given the conductivity of a Na2SO4 solution and its increase when saturated with CaSO4. Here’s a step-by-step solution:

### Step 1: Calculate the Molar Conductivity of Na2SO4
The molar conductivity (Λ) can be calculated using the formula:

\[ \Lambda = \frac{\text{Conductivity} \times 1000}{\text{Concentration}} \]

Given:
- Conductivity of Na2SO4 solution (k1) = \(2.6 \times 10^{-4} \, \text{S/cm}\)
- Concentration of Na2SO4 = \(0.001 \, \text{M}\)

Substituting the values:

\[ \Lambda_{Na2SO4} = \frac{2.6 \times 10^{-4} \times 1000}{0.001} = 260 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 2: Calculate the Molar Conductivity of SO4²⁻
The molar conductivity of Na2SO4 can be expressed as:

\[ \Lambda_{Na2SO4} = 2 \Lambda_{Na^+} + \Lambda_{SO4^{2-}} \]

Where:
- \(\Lambda_{Na^+} = 50 \, \text{S cm}^2 \text{mol}^{-1}\)

Substituting the known values:

\[ 260 = 2 \times 50 + \Lambda_{SO4^{2-}} \]

Solving for \(\Lambda_{SO4^{2-}}\):

\[ \Lambda_{SO4^{2-}} = 260 - 100 = 160 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate the Increase in Conductivity Due to CaSO4
The conductivity of the solution increases to \(k2 = 7.0 \times 10^{-4} \, \text{S/cm}\). The conductivity contributed by CaSO4 can be calculated as:

\[ \Delta k = k2 - k1 = 7.0 \times 10^{-4} - 2.6 \times 10^{-4} = 4.4 \times 10^{-4} \, \text{S/cm} \]

### Step 4: Calculate the Molar Conductivity of CaSO4
The molar conductivity for CaSO4 can be expressed as:

\[ \Lambda_{CaSO4} = \Lambda_{Ca^{2+}} + \Lambda_{SO4^{2-}} \]

Where:
- \(\Lambda_{Ca^{2+}} = 120 \, \text{S cm}^2 \text{mol}^{-1}\)

Substituting the known values:

\[ \Lambda_{CaSO4} = 120 + 160 = 280 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 5: Calculate the Concentration of Ca²⁺ from CaSO4
Using the molar conductivity of CaSO4, we can find the concentration of Ca²⁺:

\[ \text{Concentration} = \frac{\text{Conductivity} \times 1000}{\Lambda_{CaSO4}} \]

Substituting the values:

\[ \text{Concentration}_{CaSO4} = \frac{4.4 \times 10^{-4} \times 1000}{280} = 1.57 \times 10^{-3} \, \text{M} \]

### Step 6: Calculate the Solubility Product (Ksp)
The solubility product (Ksp) for CaSO4 is given by:

\[ Ksp = [Ca^{2+}][SO4^{2-}] \]

Since the solution was already saturated with Na2SO4 at \(0.001 \, \text{M}\), the concentration of SO4²⁻ will be:

\[ [SO4^{2-}] = 1.57 \times 10^{-3} + 0.001 = 2.57 \times 10^{-3} \, \text{M} \]

Now substituting the concentrations into the Ksp expression:

\[ Ksp = (1.57 \times 10^{-3})(2.57 \times 10^{-3}) = 4.04 \times 10^{-6} \]

### Final Answer
Thus, the solubility product \(Ksp\) of CaSO4 is approximately \(4.04 \times 10^{-6}\).

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