Question

\(\mathrm{K}_{s}\) of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10 .}\) If the ionic conductances of \(\mathrm{Ba}^{+}\)and \(\mathrm{SO}_{4}^{2-\text { ions are }} 64\) and \(80 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) 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) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

Short answer

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

Step-by-step solution

Understand and Write the Dissociation Reaction

Barium sulfate (\(\mathrm{BaSO}_4\)) dissociates in water according to the equation: \[ \mathrm{BaSO}_4 (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + \mathrm{SO}_4^{2-} (aq) \]The solubility product \(K_s\) of \(1 \times 10^{-10}\) represents the equilibrium concentrations of the ions.

Express Solubility Product

Given the solubility product \(K_s = 1 \times 10^{-10}\), it can be expressed as: \[K_s = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}]\]At equilibrium, \([\mathrm{Ba}^{2+}] = [\mathrm{SO}_4^{2-}] = s\) where \(s\) is the solubility.

Solve for Solubility

Since \([\mathrm{Ba}^{2+}] = [\mathrm{SO}_4^{2-}] = s\), we have: \[s^2 = 1 \times 10^{-10}\]Thus, \(s = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5}\, \text{mol/L}\).

Calculate Ionic Conductance Contributions

Ionic conductance contributions are given by: \[\lambda_{\mathrm{Ba}^{2+}} = 64\, \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\] \[\lambda_{\mathrm{SO}_4^{2-}} = 80\, \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\] Multiply each by the solubility to find specific conductance contributions.

Calculate Specific Conductance

The specific conductance \(\kappa\) is the sum of ion conductances multiplied by their concentrations:\[\kappa = (\lambda_{\mathrm{Ba}^{2+}} \times s) + (\lambda_{\mathrm{SO}_4^{2-}} \times s)\]Substitute values:\[\kappa = (64 \times 1 \times 10^{-5}) + (80 \times 1 \times 10^{-5})\]\[\kappa = 1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\]

Match with Given Options

Compare the calculated specific conductance with the provided options to find the correct answer:(a) \(1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) matches our calculation.

Key concepts

Dissociation Reaction

When discussing the dissociation reaction, it's important to understand that this is a process where a compound splits into its individual ions when dissolved in water. For barium sulfate (\(\mathrm{BaSO}_4\)), the dissociation process can be represented as follows:\[\mathrm{BaSO}_4(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + \mathrm{SO}_4^{2-}(aq)\]This balanced reaction indicates that when \(\mathrm{BaSO}_4\) dissolves, it separates into one barium ion (\(\mathrm{Ba}^{2+}\)) and one sulfate ion (\(\mathrm{SO}_4^{2-}\)). This reversible reaction reaches a state of balance or equilibrium where the rates of dissolution and precipitation become equal. The extent to which \(\mathrm{BaSO}_4\) dissolves is quantified by its solubility product (\(K_s\)), which helps predict the concentrations of these ions at equilibrium.

Ionic Conductance

Ionic conductance is a measure of how well ions can move through a solution to conduct electricity. Each type of ion, based on its size and charge, conducts electricity differently. For our discussion:

• The ionic conductance of \(\mathrm{Ba}^{2+}\) is given as 64 ohm\(^{-1}\) cm\(^{2}\) mol\(^{-1}\), indicating it conducts electricity quite well.
• Similarly, the sulfate ion (\(\mathrm{SO}_4^{2-}\)) contributes an ionic conductance of 80 ohm\(^{-1}\) cm\(^{2}\) mol\(^{-1}\).

These values are essential to calculate the solution's ability to conduct electricity when these ions are present. The higher the ionic conductance value, the better the ions are at moving and conducting within the solution.

Specific Conductance

Specific conductance, also known as conductivity, is a measure of a solution's ability to conduct electricity, dependent on both ion concentration and types of ions present.
For barium sulfate, knowing its solubility (\(s = 1 \times 10^{-5}\) mol/L) and the given ionic conductances, the specific conductance (\(\kappa\)) of the solution can be calculated using the formula:\[\kappa = (\lambda_{\mathrm{Ba}^{2+}} \times s) + (\lambda_{\mathrm{SO}_4^{2-}} \times s)\]It translates to:

• Substitute the given conductance values: \(64 \times 1 \times 10^{-5}\) for \(\mathrm{Ba}^{2+}\) and \(80 \times 1 \times 10^{-5}\) for \(\mathrm{SO}_4^{2-}\).
• This results in a specific conductance of:\[\kappa = 1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\]

This value helps determine the efficiency of the solution to carry electric current, critical for various applications in chemistry and industry.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products in a chemical reaction when it has reached a steady state. For the dissociation of \(\mathrm{BaSO}_4\), the equilibrium is described by its solubility product, \(K_s = 1 \times 10^{-10}\).
At equilibrium, the concentrations of \(\mathrm{Ba}^{2+}\) and \(\mathrm{SO}_4^{2-}\) ions are equal and can be expressed as:\[[\mathrm{Ba}^{2+}] = [\mathrm{SO}_4^{2-}] = s\]where \(s\) is the solubility of \(\mathrm{BaSO}_4\). Solving \(s^2 = 1 \times 10^{-10}\) gives \(s = 1 \times 10^{-5}\) mol/L, which represents these ion concentrations at equilibrium. Understanding these concentrations allows chemists to predict the behavior of ions in solution, crucial for controlling reaction conditions and outcomes.