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Chapter 6: Problem 80

Ksp of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10} .\) If the ionic conductances of \(\mathrm{Ba}^{++}\)and \(\mathrm{SO}_{4}\) 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

Expert verified
The specific conductance is closest to option a: \(1.44 \times 10^{-8} \text{ ohm}^{-1} \text{cm}^{-1}\).

Step by step solution

01

Determine Concentrations from Ksp

The solubility product, \( K_{sp} \), for \( \text{BaSO}_4 \) is given by the expression \( K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \). Given \( K_{sp} = 1 \times 10^{-10} \), we can determine that \([\text{Ba}^{2+}] = x\) and \([\text{SO}_4^{2-}] = x\) because they are formed in a 1:1 ratio. Hence, \( x^2 = 1 \times 10^{-10} \), resulting in \( x = \sqrt{1 \times 10^{-10}} = 10^{-5} \).
02

Calculate Ionic Conductivity Contributions

The specific conductance (\( \kappa \)) is calculated using the relation \( \kappa = c \times (\lambda_{\text{Ba}^{2+}} + \lambda_{\text{SO}_4^{2-}}) \), where \( c \) is the concentration and \( \lambda \) are the ionic molar conductivities. Given \( c = 10^{-5} \) mol/L (from Step 1), \( \lambda_{\text{Ba}^{2+}} = 64 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \), and \( \lambda_{\text{SO}_4^{2-}} = 80 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \), we have: \[\kappa = 10^{-5} \times (64 + 80) = 10^{-5} \times 144 = 1.44 \times 10^{-3} \text{ ohm}^{-1} \text{cm}^{-1}.\]
03

Adjust for Unit Conversion

To express the answer in terms of the provided options, we consider the units. Our calculation yielded \( 1.44 \times 10^{-3} \text{ ohm}^{-1} \text{cm}^{-1} \), which can be rewritten as \( 144 \times 10^{-5} \text{ ohm}^{-1} \text{cm}^{-1} \) by adjusting the placement of the decimal (i.e., multiplying by 100). Thus, converting further, \( 144 \times 10^{-5} \text{ ohm}^{-1} \text{cm}^{-1} = 1.44 \times 10^{-3} \text{ ohm}^{-1} \text{cm}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ionic Conductance
Ionic conductance is a measure of how well an ion can carry an electric charge through a solution. It is denoted by \( \lambda \) and is expressed in units of \( \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \). The value of ionic conductance reveals how effective a given ion is at conducting electricity in a solution.

In the case of barium sulfate \( (\text{BaSO}_4) \), the ionic conductance values for \( \text{Ba}^{++} \) and \( \text{SO}_4^{2-} \) are important to understand how they contribute to the overall conductance of the solution. Ionic conductance depends on the mobility of the ions: ions with higher mobility (i.e., smaller size) or charge tend to have higher conductance values.

Here, the values are given as 64 and 80 \( \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \) for \( \text{Ba}^{++} \) and \( \text{SO}_4^{2-} \), respectively. This suggests that the sulfate ion is more efficient in conducting electricity compared to the barium ion.
Exploring Specific Conductance
Specific conductance, also known as conductivity (\( \kappa \)), is a measure of a solution鈥檚 ability to conduct electricity. It is expressed in units of \( \text{ohm}^{-1} \text{cm}^{-1} \). It differs from ionic conductance in that it considers the entire solution, not just a single ion.

To calculate specific conductance, one must consider both the concentration of ions and their individual ionic conductivities. The specific conductance is calculated using the formula:
  • \( \kappa = c \times (\lambda_{\text{Ba}^{2+}} + \lambda_{\text{SO}_4^{2-}}) \)
where \( c \) is the molar concentration of the ions. After calculating, the solution yields specific conductance, giving an indication of how well the entire solution can conduct electricity. In this context, the concentration calculated from the solubility product \( (K_{sp}) \) was used to find that the specific conductance of the \( \text{BaSO}_4 \) solution is \( 1.44 \times 10^{-3} \text{ ohm}^{-1} \text{cm}^{-1} \).
Delving into Ionic Molar Conductivity
Ionic molar conductivity is a concept closely linked to ionic conductance. It describes how efficiently ions at a particular concentration conduct electricity. Molar conductivity is given by the formula:
  • \( \Lambda_m = \frac{\kappa}{c} \)
This formula allows concentration effects to be normalized, making it easier to compare conductivity across different solutions with varying concentrations.

The units of molar conductivity are \( \text{ohm}^{-1}\text{cm}^2\text{mol}^{-1} \), blending the broader measure of solution conductivity with concentration factors. As the concentration of a solution decreases, the molar conductivity typically increases owing to reduced interionic attractions, often leading ions to conduct electricity more freely. Knowing the molar conductivity is fundamental when interpreting how changes in concentration affect a solution's ability to conduct electricity.

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