301. For a weak electrolyte, which expression correctly gives the degree of dissociation from conductivity data?
ⓐ. \(\alpha = \frac{\Lambda_m^\circ}{\Lambda_m}\)
ⓑ. \(\alpha = \frac{\kappa}{\Lambda_m^\circ}\)
ⓒ. \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\)
ⓓ. \(\alpha = \frac{\kappa}{c}\)
Correct Answer: \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\)
Explanation: Degree of dissociation for a weak electrolyte is obtained by comparing the molar conductivity at the given concentration with the limiting molar conductivity. The correct relation is
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\].
This expression is specifically useful for weak electrolytes.
302. A weak acid solution has \(\Lambda_m = 5.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\) and \(\Lambda_m^\circ = 25.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\). What is its percentage dissociation?
ⓐ. \(20%\)
ⓑ. \(5%\)
ⓒ. \(25%\)
ⓓ. \(50%\)
Correct Answer: \(20%\)
Explanation: \(\textbf{Given:}\)
\[\Lambda_m = 5.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\Lambda_m^\circ = 25.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Percentage dissociation
\(\textbf{Relevant formula:}\)
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
\(\textbf{Why this formula applies:}\)
For a weak electrolyte, degree of dissociation is obtained from conductivity data.
\(\textbf{Substitution:}\)
\[\alpha = \frac{5.0 \times 10^{-3}}{25.0 \times 10^{-3}}\]
\(\textbf{Intermediate simplification:}\)
The factor \(10^{-3}\) cancels.
So,
\[\alpha = \frac{5.0}{25.0} = 0.20\]
\(\textbf{Convert to percentage:}\)
Percentage dissociation \(= 0.20 \times 100 = 20%\)
\(\textbf{Final Answer:}\)
Percentage dissociation \(= 20%\)
303. The conductivity of a saturated solution of a sparingly soluble \(1:1\) electrolyte is \(3.0 \times 10^{-4}\,\text{S m}^{-1}\). Its limiting molar conductivity is \(1.5 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\). What is its solubility in \(\text{mol m}^{-3}\)?
ⓐ. \(2.0 \times 10^{-3}\,\text{mol m}^{-3}\)
ⓑ. \(3.0 \times 10^{-4}\,\text{mol m}^{-3}\)
ⓒ. \(2.0 \times 10^{-1}\,\text{mol m}^{-3}\)
ⓓ. \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
Correct Answer: \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
Explanation: \(\textbf{Given:}\)
\[\kappa = 3.0 \times 10^{-4}\,\text{S m}^{-1}\]
\[\Lambda_m^\circ = 1.5 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Solubility, \(c\), in \(\text{mol m}^{-3}\)
\(\textbf{Relevant formula:}\)
\[\Lambda_m = \frac{\kappa}{c}\]
\(\textbf{Why this formula applies:}\)
For a very dilute saturated solution of a sparingly soluble salt, \(\Lambda_m \approx \Lambda_m^\circ\).
So,
\[c = \frac{\kappa}{\Lambda_m^\circ}\]
\(\textbf{Substitution:}\)
\[c = \frac{3.0 \times 10^{-4}}{1.5 \times 10^{-2}}\]
\(\textbf{Intermediate simplification:}\)
\[c = \frac{3.0}{1.5} \times 10^{-4+2}\]
\[c = 2.0 \times 10^{-2}\]
\(\textbf{Unit check:}\)
\[\frac{\text{S m}^{-1}}{\text{S m}^2\,\text{mol}^{-1}} = \text{mol m}^{-3}\]
\(\textbf{Final Answer:}\)
\[c = 2.0 \times 10^{-2}\,\text{mol m}^{-3}\]
304. A weak acid of concentration \(0.010\,\text{mol L}^{-1}\) has \(\Lambda_m = 3.9 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\) and \(\Lambda_m^\circ = 39.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\). What is its dissociation constant \(K_a\)?
ⓐ. \(1.1 \times 10^{-3}\)
ⓑ. \(1.1 \times 10^{-4}\)
ⓒ. \(9.0 \times 10^{-3}\)
ⓓ. \(9.0 \times 10^{-5}\)
Correct Answer: \(1.1 \times 10^{-4}\)
Explanation: \(\textbf{Given:}\)
\[c = 0.010\,\text{mol L}^{-1}\]
\[\Lambda_m = 3.9 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\Lambda_m^\circ = 39.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Dissociation constant, \(K_a\)
\(\textbf{Relevant relations:}\)
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
For a weak acid,
\[K_a = \frac{c\alpha^2}{1-\alpha}\]
\(\textbf{Why these formulas apply:}\)
The degree of dissociation is first found from conductivity data, and then used in the weak-acid dissociation expression.
Find \(\alpha\):
\[\alpha = \frac{3.9 \times 10^{-3}}{39.0 \times 10^{-3}} = 0.10\]
Substitution into the \(K_a\) expression:
\[K_a = \frac{0.010 \times (0.10)^2}{1 - 0.10}\]
\(\textbf{Intermediate simplification:}\)
\[(0.10)^2 = 0.01\]
So,
\[K_a = \frac{0.010 \times 0.01}{0.90}\]
\[K_a = \frac{1.0 \times 10^{-4}}{0.90}\]
\(\textbf{Final simplification:}\)
\[K_a \approx 1.1 \times 10^{-4}\]
\(\textbf{Final Answer:}\)
\[K_a = 1.1 \times 10^{-4}\]
305. Which statement is correct for a weak electrolyte when dilution increases?
ⓐ. Its degree of dissociation decreases because ions become farther apart.
ⓑ. Its degree of dissociation increases, so molar conductivity rises sharply.
ⓒ. Its conductivity increases because the number of ions per unit volume increases.
ⓓ. Its limiting molar conductivity becomes zero.
Correct Answer: Its degree of dissociation increases, so molar conductivity rises sharply.
Explanation: A weak electrolyte is only partially ionized in solution. On dilution, ionization increases noticeably, producing a larger fraction of ions from the same amount of electrolyte. This causes molar conductivity to rise sharply, even though conductivity itself usually decreases.
306. A weak electrolyte has \(\Lambda_m = 2.4 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\) at a certain concentration and \(\Lambda_m^\circ = 24.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\). Its degree of dissociation is
ⓐ. \(0.01\)
ⓑ. \(0.10\)
ⓒ. \(0.20\)
ⓓ. \(0.50\)
Correct Answer: \(0.10\)
Explanation: \(\textbf{Given:}\)
\[\Lambda_m = 2.4 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\Lambda_m^\circ = 24.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Degree of dissociation, \(\alpha\)
\(\textbf{Relevant formula:}\)
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
\(\textbf{Why this formula applies:}\)
For a weak electrolyte, the fraction dissociated is obtained from the ratio of the molar conductivity at the given concentration to the limiting molar conductivity.
\(\textbf{Substitution:}\)
\[\alpha = \frac{2.4 \times 10^{-3}}{24.0 \times 10^{-3}}\]
\(\textbf{Intermediate simplification:}\)
The factor \(10^{-3}\) cancels.
So,
\[\alpha = \frac{2.4}{24.0}\]
\(\textbf{Final simplification:}\)
\[\alpha = 0.10\]
\(\textbf{Final Answer:}\)
\[\alpha = 0.10\]
307. A weak acid solution has concentration \(0.020\,\text{mol L}^{-1}\) and degree of dissociation \(\alpha = 0.05\). Its dissociation constant is
ⓐ. \(5.0 \times 10^{-3}\)
ⓑ. \(5.3 \times 10^{-4}\)
ⓒ. \(5.3 \times 10^{-5}\)
ⓓ. \(2.5 \times 10^{-3}\)
Correct Answer: \(5.3 \times 10^{-5}\)
Explanation: \(\textbf{Given:}\)
\[c = 0.020\,\text{mol L}^{-1}\]
\[\alpha = 0.05\]
\(\textbf{Required:}\)
Dissociation constant, \(K_a\)
\(\textbf{Relevant formula:}\)
For a weak acid,
\[K_a = \frac{c\alpha^2}{1-\alpha}\]
\(\textbf{Why this formula applies:}\)
A weak acid is only partially dissociated, so the equilibrium expression is written in terms of degree of dissociation.
\(\textbf{Substitution:}\)
\[K_a = \frac{0.020 \times (0.05)^2}{1 - 0.05}\]
\(\textbf{Intermediate simplification:}\)
\[(0.05)^2 = 0.0025\]
So,
\[K_a = \frac{0.020 \times 0.0025}{0.95}\]
\[K_a = \frac{5.0 \times 10^{-5}}{0.95}\]
\(\textbf{Final simplification:}\)
\[K_a \approx 5.3 \times 10^{-5}\]
\(\textbf{Final Answer:}\)
\[K_a = 5.3 \times 10^{-5}\]
308. A saturated solution of a sparingly soluble \(1:1\) electrolyte has conductivity \(4.5 \times 10^{-4}\,\text{S m}^{-1}\). If its limiting molar conductivity is \(2.25 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\), the solubility is
ⓐ. \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
ⓑ. \(5.0 \times 10^{-2}\,\text{mol m}^{-3}\)
ⓒ. \(1.0 \times 10^{-2}\,\text{mol m}^{-3}\)
ⓓ. \(2.0 \times 10^{-3}\,\text{mol m}^{-3}\)
Correct Answer: \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
Explanation: \(\textbf{Given:}\)
\[\kappa = 4.5 \times 10^{-4}\,\text{S m}^{-1}\]
\[\Lambda_m^\circ = 2.25 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Solubility, \(c\)
\(\textbf{Relevant formula:}\)
For a very dilute saturated solution of a sparingly soluble electrolyte,
\[c = \frac{\kappa}{\Lambda_m^\circ}\]
\(\textbf{Why this formula applies:}\)
At very low concentration, the molar conductivity is taken approximately equal to the limiting molar conductivity.
\(\textbf{Substitution:}\)
\[c = \frac{4.5 \times 10^{-4}}{2.25 \times 10^{-2}}\]
\(\textbf{Intermediate simplification:}\)
\[c = \frac{4.5}{2.25} \times 10^{-4+2}\]
\[c = 2.0 \times 10^{-2}\]
\(\textbf{Unit check:}\)
\[\frac{\text{S m}^{-1}}{\text{S m}^2\,\text{mol}^{-1}} = \text{mol m}^{-3}\]
\(\textbf{Final Answer:}\)
\[c = 2.0 \times 10^{-2}\,\text{mol m}^{-3}\]
309. Which statement about the use of conductivity data for weak electrolytes is correct?
ⓐ. Conductivity data cannot be used to obtain any equilibrium information.
ⓑ. Degree of dissociation is found from \(\alpha = \frac{\Lambda_m^\circ}{\Lambda_m}\).
ⓒ. Limiting molar conductivity of a weak electrolyte is usually obtained most directly by simple linear extrapolation of its own data.
ⓓ. Conductivity data can be used to estimate degree of dissociation and then the dissociation constant.
Correct Answer: Conductivity data can be used to estimate degree of dissociation and then the dissociation constant.
Explanation: For a weak electrolyte, one first obtains \(\Lambda_m^\circ\), usually through Kohlrausch’s law. Then \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\) gives the degree of dissociation. This value can be used to calculate the dissociation constant.
310. The limiting ionic conductivities are:
\[\lambda^\circ(Ca^{2+}) = 11.9 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\lambda^\circ(NO_3^-) = 7.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
What is \(\Lambda_m^\circ\left(Ca(NO_3)_2\right)\)?
ⓐ. \(19.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓑ. \(26.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓒ. \(33.2 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓓ. \(14.2 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Correct Answer: \(26.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Explanation: \(\textbf{Given:}\)
\[\lambda^\circ(Ca^{2+}) = 11.9 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\lambda^\circ(NO_3^-) = 7.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
\[\Lambda_m^\circ\left(Ca(NO_3)_2\right)\]
\(\textbf{Relevant law:}\)
Kohlrausch’s law of independent migration of ions
\(\textbf{Why this law applies:}\)
At infinite dilution, each ion contributes independently, and stoichiometric coefficients must be included.
\(\textbf{Write the expression:}\)
\[\Lambda_m^\circ\left(Ca(NO_3)_2\right) = \lambda^\circ(Ca^{2+}) + 2\lambda^\circ(NO_3^-)\]
\(\textbf{Substitution:}\)
\[\Lambda_m^\circ = 11.9 \times 10^{-3} + 2(7.1 \times 10^{-3})\]
\(\textbf{Intermediate simplification:}\)
\[2(7.1 \times 10^{-3}) = 14.2 \times 10^{-3}\]
So,
\[\Lambda_m^\circ = 11.9 \times 10^{-3} + 14.2 \times 10^{-3}\]
\(\textbf{Final simplification:}\)
\[\Lambda_m^\circ = 26.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Final Answer:}\)
\[\Lambda_m^\circ\left(Ca(NO_3)_2\right) = 26.1 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
311. Assertion: For a weak electrolyte, molar conductivity increases sharply on dilution.
Reason: Dilution increases its degree of ionization significantly.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Weak electrolytes are only partially ionized in solution. On dilution, their degree of ionization rises significantly, producing more ions and increasing their conducting ability per mole. Therefore the sharp increase in molar conductivity is correctly explained by the increase in ionization.
312. A weak electrolyte has \(\alpha = 0.20\) at a certain concentration and \(\Lambda_m^\circ = 18.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\). What is its molar conductivity at that concentration?
ⓐ. \(9.0 \times 10^{-4}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓑ. \(3.6 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓒ. \(1.44 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓓ. \(9.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Correct Answer: \(3.6 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Explanation: \(\textbf{Given:}\)
\[\alpha = 0.20\]
\[\Lambda_m^\circ = 18.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
\[\Lambda_m\]
\(\textbf{Relevant formula:}\)
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
\(\textbf{Why this formula applies:}\)
For a weak electrolyte, degree of dissociation is directly related to molar conductivity.
\(\textbf{Rearrangement:}\)
\[\Lambda_m = \alpha \Lambda_m^\circ\]
\(\textbf{Substitution:}\)
\[\Lambda_m = 0.20 \times 18.0 \times 10^{-3}\]
\(\textbf{Intermediate simplification:}\)
\[\Lambda_m = 3.6 \times 10^{-3}\]
\(\textbf{Unit check:}\)
The unit remains \(\text{S m}^2\,\text{mol}^{-1}\).
\(\textbf{Final Answer:}\)
\[\Lambda_m = 3.6 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
313. Which method is commonly used to obtain the limiting molar conductivity of a strong electrolyte?
ⓐ. Direct use of \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\) at any concentration
ⓑ. Extrapolation of the plot of \(\Lambda_m\) versus \(\sqrt{c}\) to zero concentration
ⓒ. Measuring conductivity only at the highest concentration available
ⓓ. Subtracting resistance of the solvent from the cell emf
Correct Answer: Extrapolation of the plot of \(\Lambda_m\) versus \(\sqrt{c}\) to zero concentration
Explanation: Strong electrolytes are almost completely ionized even at ordinary concentrations. Their molar conductivity changes only slightly with dilution, and the variation with \(\sqrt{c}\) is nearly linear. So the plot of \(\Lambda_m\) against \(\sqrt{c}\) can be extrapolated to \(\sqrt{c}=0\) to obtain \(\Lambda_m^\circ\).
314. The limiting ionic conductivities are:
\[\lambda^\circ(Na^+) = 5.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\lambda^\circ(SO_4^{2-}) = 16.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
What is \(\Lambda_m^\circ(Na_2SO_4)\)?
ⓐ. \(11.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓑ. \(21.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓒ. \(26.0 \times 10^{-6}\,\text{S m}^2\,\text{mol}^{-1}\)
ⓓ. \(26.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Correct Answer: \(26.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\)
Explanation: \(\textbf{Given:}\)
\[\lambda^\circ(Na^+) = 5.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\lambda^\circ(SO_4^{2-}) = 16.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
\[\Lambda_m^\circ(Na_2SO_4)\]
\(\textbf{Relevant law:}\)
Kohlrausch’s law of independent migration of ions
\(\textbf{Why this law applies:}\)
At infinite dilution, the limiting molar conductivity is the sum of ionic contributions with proper stoichiometric coefficients.
\(\textbf{Write the expression:}\)
\[\Lambda_m^\circ(Na_2SO_4) = 2\lambda^\circ(Na^+) + \lambda^\circ(SO_4^{2-})\]
\(\textbf{Substitution:}\)
\[\Lambda_m^\circ = 2(5.0 \times 10^{-3}) + 16.0 \times 10^{-3}\]
\(\textbf{Intermediate simplification:}\)
\[2(5.0 \times 10^{-3}) = 10.0 \times 10^{-3}\]
So,
\[\Lambda_m^\circ = 10.0 \times 10^{-3} + 16.0 \times 10^{-3}\]
\(\textbf{Final simplification:}\)
\[\Lambda_m^\circ = 26.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Final Answer:}\)
\[\Lambda_m^\circ(Na_2SO_4) = 26.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
315. Why does the molar conductivity of a strong electrolyte increase only slightly on dilution?
ⓐ. Because it is already highly ionized
ⓑ. Because its degree of ionization falls on dilution
ⓒ. Because the number of ions per unit volume rises sharply on dilution
ⓓ. Because strong electrolytes do not obey any conductivity relation
Correct Answer: Because it is already highly ionized
Explanation: Strong electrolytes are almost fully dissociated even before large dilution. Therefore dilution does not create many additional ions. The increase in molar conductivity mainly comes from reduced interionic attraction and better mobility, so the rise is only slight.
316. A weak acid of concentration \(0.010\,\text{mol L}^{-1}\) has \(\Lambda_m = 2.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\) and \(\Lambda_m^\circ = 20.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\). What is its dissociation constant?
ⓐ. \(1.0 \times 10^{-2}\)
ⓑ. \(9.0 \times 10^{-4}\)
ⓒ. \(1.1 \times 10^{-4}\)
ⓓ. \(1.0 \times 10^{-5}\)
Correct Answer: \(1.1 \times 10^{-4}\)
Explanation: \(\textbf{Given:}\)
\[c = 0.010\,\text{mol L}^{-1}\]
\[\Lambda_m = 2.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\[\Lambda_m^\circ = 20.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Dissociation constant, \(K_a\)
\(\textbf{Relevant relations:}\)
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
\[K_a = \frac{c\alpha^2}{1-\alpha}\]
\(\textbf{Why these formulas apply:}\)
For a weak acid, conductivity data first gives the degree of dissociation, which is then used in the equilibrium expression.
Find \(\alpha\):
\[\alpha = \frac{2.0 \times 10^{-3}}{20.0 \times 10^{-3}} = 0.10\]
\(\textbf{Substitute into the dissociation-constant formula:}\)
\[K_a = \frac{0.010(0.10)^2}{1-0.10}\]
\(\textbf{Intermediate simplification:}\)
\[(0.10)^2 = 0.01\]
So,
\[K_a = \frac{0.010 \times 0.01}{0.90}\]
\[K_a = \frac{1.0 \times 10^{-4}}{0.90}\]
\(\textbf{Final simplification:}\)
\[K_a \approx 1.1 \times 10^{-4}\]
\(\textbf{Final Answer:}\)
\[K_a = 1.1 \times 10^{-4}\]
317. The conductivity of a saturated solution of a sparingly soluble \(1:1\) electrolyte is \(6.0 \times 10^{-4}\,\text{S m}^{-1}\). Its limiting molar conductivity is \(3.0 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\). What is its solubility?
ⓐ. \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
ⓑ. \(2.0 \times 10^{-3}\,\text{mol m}^{-3}\)
ⓒ. \(5.0 \times 10^{-2}\,\text{mol m}^{-3}\)
ⓓ. \(5.0 \times 10^{-3}\,\text{mol m}^{-3}\)
Correct Answer: \(2.0 \times 10^{-2}\,\text{mol m}^{-3}\)
Explanation: \(\textbf{Given:}\)
\[\kappa = 6.0 \times 10^{-4}\,\text{S m}^{-1}\]
\[\Lambda_m^\circ = 3.0 \times 10^{-2}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Solubility, \(c\)
\(\textbf{Relevant formula:}\)
\[c = \frac{\kappa}{\Lambda_m^\circ}\]
\(\textbf{Why this formula applies:}\)
For a very dilute saturated solution of a sparingly soluble electrolyte, \(\Lambda_m \approx \Lambda_m^\circ\).
\(\textbf{Substitution:}\)
\[c = \frac{6.0 \times 10^{-4}}{3.0 \times 10^{-2}}\]
\(\textbf{Intermediate simplification:}\)
\[c = \frac{6.0}{3.0} \times 10^{-4+2}\]
\[c = 2.0 \times 10^{-2}\]
\(\textbf{Unit check:}\)
\[\frac{\text{S m}^{-1}}{\text{S m}^2\,\text{mol}^{-1}} = \text{mol m}^{-3}\]
\(\textbf{Final Answer:}\)
\[c = 2.0 \times 10^{-2}\,\text{mol m}^{-3}\]
318. Which statement is correct about the conductivity method for weak electrolytes?
ⓐ. It gives degree of dissociation directly from resistance alone without using \(\Lambda_m^\circ\)
ⓑ. It cannot be used for weak acids or weak bases
ⓒ. It applies only to electrolytes that are fully ionized at all concentrations
ⓓ. It uses \(\Lambda_m^\circ\) to estimate \(\alpha\), and then equilibrium information can be obtained
Correct Answer: It uses \(\Lambda_m^\circ\) to estimate \(\alpha\), and then equilibrium information can be obtained
Explanation: For a weak electrolyte, the degree of dissociation is found from \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\). Once \(\alpha\) is known, the dissociation constant can be calculated using the concentration of the solution. This makes conductivity data very useful for weak electrolytes.
319. A weak electrolyte solution has conductivity \(\kappa = 0.24\,\text{S m}^{-1}\) and concentration \(c = 40\,\text{mol m}^{-3}\). If \(\Lambda_m^\circ = 15.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\), what is its degree of dissociation?
ⓐ. \(0.60\)
ⓑ. \(0.40\)
ⓒ. \(1.60\)
ⓓ. \(0.16\)
Correct Answer: \(0.40\)
Explanation: \(\textbf{Given:}\)
\[\kappa = 0.24\,\text{S m}^{-1}\]
\[c = 40\,\text{mol m}^{-3}\]
\[\Lambda_m^\circ = 15.0 \times 10^{-3}\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Required:}\)
Degree of dissociation, \(\alpha\)
\(\textbf{Relevant formulas:}\)
\[\Lambda_m = \frac{\kappa}{c}\]
\[\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\]
\(\textbf{Why these formulas apply:}\)
First the molar conductivity is found from conductivity and concentration. Then degree of dissociation is obtained by comparing with the limiting molar conductivity.
Find \(\Lambda_m\):
\[\Lambda_m = \frac{0.24}{40} = 0.006\,\text{S m}^2\,\text{mol}^{-1}\]
Write \(\Lambda_m^\circ\) in decimal form:
\[\Lambda_m^\circ = 15.0 \times 10^{-3} = 0.015\,\text{S m}^2\,\text{mol}^{-1}\]
\(\textbf{Substitute into the dissociation formula:}\)
\[\alpha = \frac{0.006}{0.015}\]
\(\textbf{Final simplification:}\)
\[\alpha = 0.40\]
\(\textbf{Final Answer:}\)
\[\alpha = 0.40\]
320. Assertion: Conductivity decreases on dilution, but molar conductivity increases on dilution.
Reason: Dilution reduces the number of ions per unit volume, while also reducing interionic attraction.
ⓐ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓑ. Assertion is true, but Reason is false.
ⓒ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Conductivity depends on the number of ions present per unit volume, so it decreases on dilution. Molar conductivity refers to the conducting ability of one mole of electrolyte and increases because ions move more freely when interionic attraction decreases. Thus the Reason correctly explains the Assertion.
