At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:
The conductivity of a saturated aqueous solution of AgCl is $1.40 \times 10^{-6}$ S cm\(^{-1}\) at 300 K. If the solubility of AgCl in water at 300 K is X mol L\(^{-1}\), then log$_{10}$(X\(^{-1}\)) is
(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)}
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For sparingly soluble salts, solubility is equal to the concentration in a saturated solution. The molar conductivity for such a solution is typically assumed to be equal to its limiting molar conductivity. Kohlrausch's law is essential for obtaining limiting conductivities of weak or sparingly soluble electrolytes from strong ones.
Step 1: Understanding the Question:
The problem involves calculating the solubility (X) of a sparingly soluble salt (AgCl) using conductivity data. We need to find the limiting molar conductivity of AgCl, then use it along with the given conductivity of its saturated solution to find X, and finally calculate log$_{10}$(X\(^{-1}\)). Step 2: Key Formula or Approach:
1. Kohlrausch's Law: To find the limiting molar conductivity of AgCl (\(\Lambda_m^0(AgCl)\)), we can use the limiting molar conductivities of strong electrolytes:
\[ \Lambda_m^0(AgCl) = \Lambda_m^0(AgNO_3) + \Lambda_m^0(NaCl) - \Lambda_m^0(NaNO_3) \]
2. Relation between Conductivity and Molar Conductivity: For a saturated solution of a sparingly soluble salt, its molar conductivity is given by:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
Where \(\kappa\) is conductivity in S cm\(^{-1}\), \(C\) is concentration in mol L\(^{-1}\) (which is solubility X in this case), and \(\Lambda_m\) is in S cm\(^2\) mol\(^{-1}\).
3. Given that \(\Lambda_m\) for saturated AgCl is equal to its limiting molar conductivity (\(\Lambda_m^0\)), we can write:
\[ \Lambda_m^0(AgCl) = \frac{\kappa(AgCl) \times 1000}{X} \] Step 3: Detailed Explanation: Part 1: Determine limiting molar conductivities from the table.
For strong electrolytes, molar conductivity decreases slightly with increasing concentration. To get the limiting molar conductivity (\(\Lambda_m^0\)), we generally take the value at the lowest given concentration or extrapolate to zero concentration. Since the decrease is small, we can use the value at 0.01 M as a good approximation for infinite dilution.
- $\Lambda_m^0(NaNO_3) \approx 111 \text{ S cm}^2 \text{ mol}^{-1}$ (at 0.01 M)
- $\Lambda_m^0(NaCl) \approx 117 \text{ S cm}^2 \text{ mol}^{-1}$ (at 0.01 M)
- $\Lambda_m^0(AgNO_3) \approx 125 \text{ S cm}^2 \text{ mol}^{-1}$ (at 0.01 M) Part 2: Calculate \(\Lambda_m^0(AgCl)\) using Kohlrausch's Law.
\[ \Lambda_m^0(AgCl) = \Lambda_m^0(AgNO_3) + \Lambda_m^0(NaCl) - \Lambda_m^0(NaNO_3) \]
\[ \Lambda_m^0(AgCl) = 125 + 117 - 111 \]
\[ \Lambda_m^0(AgCl) = 242 - 111 = 131 \text{ S cm}^2 \text{ mol}^{-1} \] Part 3: Calculate solubility X of AgCl.
Given conductivity of saturated AgCl solution, \( \kappa(AgCl) = 1.40 \times 10^{-6} \text{ S cm}^{-1} \).
Using the formula \(\Lambda_m^0(AgCl) = \frac{\kappa(AgCl) \times 1000}{X}\):
\[ 131 = \frac{1.40 \times 10^{-6} \times 1000}{X} \]
\[ 131 = \frac{1.40 \times 10^{-3}}{X} \]
\[ X = \frac{1.40 \times 10^{-3}}{131} \]
\[ X \approx 0.010687 \times 10^{-3} \text{ mol L}^{-1} \]
\[ X \approx 1.0687 \times 10^{-5} \text{ mol L}^{-1} \] Part 4: Calculate log$_{10$(X\(^{-1}\)).}
\[ X^{-1} = \frac{1}{X} = \frac{1}{1.0687 \times 10^{-5}} \]
\[ X^{-1} \approx 0.9357 \times 10^5 \]
\[ X^{-1} \approx 9.357 \times 10^4 \]
Now, calculate $\log_{10}(X^{-1})$:
\[ \log_{10}(X^{-1}) = \log_{10}(9.357 \times 10^4) \]
\[ = \log_{10}(9.357) + \log_{10}(10^4) \]
\[ \approx 0.971 + 4 = 4.971 = 5 \] Step 4: Final Answer:
Final answer rounds off to 5.
