Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{K}_{\text {p }}\) of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10 .}\) If the ionic conductances of \(\mathrm{Ba}^{+}\)and \(\mathrm{SO}_{4}^{2-}\) ions are 64 and \(80 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}\) - respectively, then its specific conductance is (a) \(1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (b) \(144 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (c) \(1.44 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (d) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

Step-by-Step Solution

Answer

The specific conductance is \(1.44 \times 10^{-8} \text{ ohm}^{-1} \text{ cm}^{-1}\).

Step 1: Identify Known Values

The problem provides the solubility product constant \( K_{sp} \) of \( BaSO_4 \), which is \( 1 \times 10^{-10} \). It also gives the ionic conductances of \( Ba^{2+} \) and \( SO_4^{2-} \), which are \( 64 \) and \( 80 \) ohm\(^{-1} \) cm\(^2\) mol\(^{-1}\), respectively.

Step 2: Use Solubility Product to Find Solubility

The solubility product \( K_{sp} \) relates to the solubility \( s \) by the equation \( K_{sp} = s^2 \) for \( BaSO_4 \), since it dissociates into \( Ba^{2+} \) and \( SO_4^{2-} \). We solve for \( s \): \[ s = \sqrt{K_{sp}} = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \].

Step 3: Calculate Molar Conductance

The molar conductance \( \Lambda_m \) can be calculated using the formula \( \Lambda_m = \lambda_{Ba^{2+}} + \lambda_{SO_4^{2-}} \), where \( \lambda \) are the ionic conductances. Substitute the values: \[ \Lambda_m = 64 + 80 = 144 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \].

Step 4: Calculate Specific Conductance

Specific conductance, \( \kappa \), is calculated as \( \kappa = \Lambda_m \cdot s \). Substitute the values for \( \Lambda_m \) and \( s \): \[ \kappa = 144 \times 10^{-5} = 1.44 \times 10^{-8} \text{ ohm}^{-1} \text{ cm}^{-1} \].

Key Concepts

Solubility Product

The solubility product, denoted as \( K_{sp} \), is a fundamental concept in chemistry, especially when dealing with sparingly soluble salts like barium sulfate (\( BaSO_4 \)).
\( K_{sp} \) is an equilibrium constant that refers to the product of the molar concentrations of the ions involved in the dissolution of a compound in water, each raised to the power of their respective stoichiometric coefficients.
For example, when \( BaSO_4 \) dissolves in water, it dissociates into \( Ba^{2+} \) and \( SO_4^{2-} \) ions:
\[ BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \]
Therefore, \( K_{sp} \) can be expressed as the product of the concentrations of these ions:
\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]
Since \( BaSO_4 \) dissociates into one \( Ba^{2+} \) ion and one \( SO_4^{2-} \) ion, the concentrations of both ions are the same, denoted as \( s \). Thus, \( K_{sp} = s^2 \).

• This means if \( K_{sp} = 1 \times 10^{-10} \), then \( s = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \) mol/L.

Molar Conductance

Molar conductance, \( \Lambda_m \), is a measure of how well a solute can conduct electricity in solution. It is particularly important when analyzing the conductance of ions in dilute solutions.
The formula for molar conductance at a given concentration is:
\[ \Lambda_m = \lambda_{+} + \lambda_{-} \]
where \( \lambda_{+} \) and \( \lambda_{-} \) represent the ionic conductances of the cation and anion, respectively.

• For \( BaSO_4 \), with \( \lambda_{Ba^{2+}} = 64 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \) and \( \lambda_{SO_4^{2-}} = 80 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \), the molar conductance is:
  \( \Lambda_m = 64 + 80 = 144 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \).

Molar conductance increases as the concentration of the solute decreases, making it more pronounced in dilute solutions where ions are more mobile.

Specific Conductance

Specific conductance, also known as conductivity, is a measure of a solution's ability to conduct electricity and is denoted by \( \kappa \). This property depends on the number of ions present and their mobility.
Specific conductance is calculated using the formula:
\[ \kappa = \Lambda_m \times c \]
where \( \Lambda_m \) is the molar conductance and \( c \) is the concentration, or solubility, of the solution.

• In the case of \( BaSO_4 \), with \( \Lambda_m = 144 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \) and solubility \( s = 1 \times 10^{-5} \text{ mol/L} \):
  \( \kappa = 144 \times 10^{-5} = 1.44 \times 10^{-8} \text{ ohm}^{-1} \text{ cm}^{-1} \).

This shows that although \( BaSO_4 \) has low solubility, its specific conductance can be calculated by taking into account both its molar conductance and its solubility.