Question

The molar conductivities at infinite dilution for Na₂SO₄, K₂SO₄, KCl, HCl, and HCOONa at 300 K are 260, 308, 150, 426, and 105 S·cm²·mol⁻¹, respectively. What will be the molar conductivity at infinite dilution (Λ⁰) for formic acid in the same unit?

• A. 480
• B. 429
• C. 430
• D. 405

Hint:

The key is to properly combine the given conductivities to isolate the needed ion contributions.

Answer 1

The Correct Option is D

The molar conductivities at infinite dilution for \(\text{Na}_2\text{SO}_4\), \(\text{K}_2\text{SO}_4\), \(\text{KCl}\), \(\text{HCl}\), and \(\text{HCOONa}\) at 300 K are 260, 308, 150, 426, and 105 S\(\cdot\)cm\(^2\)\(\cdot\)mol\(^{-1}\), respectively. We need to calculate the molar conductivity at infinite dilution (\(\Lambda^0\)) for formic acid (HCOOH) in the same units. Using Kohlrausch’s law, the molar conductivity at infinite dilution of an electrolyte is the sum of the ionic molar conductivities of its ions, weighted by their stoichiometric coefficients: \[\begin{equation} \Lambda^0_{\text{electrolyte}} = \nu_+ \lambda^0_+ + \nu_- \lambda^0_- \end{equation}\] where \(\nu_+\) and \(\nu_-\) are the number of cations and anions per formula unit, and \(\lambda^0_+\) and \(\lambda^0_-\) are the ionic molar conductivities at infinite dilution. Formic acid (HCOOH) dissociates as: \[ \text{HCOOH} \rightarrow \text{H}^+ + \text{HCOO}^- \] Thus, \(\nu_+ = 1\) (for \(\text{H}^+\)), \(\nu_- = 1\) (for \(\text{HCOO}^-\)), and: \[\begin{equation} \Lambda^0_{\text{HCOOH}} = \lambda^0_{\text{H}^+} + \lambda^0_{\text{HCOO}^-} \end{equation}\] We are given:

• \(\Lambda^0_{\text{Na}_2\text{SO}_4} = 260 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}\)
• \(\Lambda^0_{\text{K}_2\text{SO}_4} = 308 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}\)
• \(\Lambda^0_{\text{KCl}} = 150 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}\)
• \(\Lambda^0_{\text{HCl}} = 426 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}\)
• \(\Lambda^0_{\text{HCOONa}} = 105 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}\)

Our goal is to find \(\lambda^0_{\text{H}^+}\) and \(\lambda^0_{\text{HCOO}^-}\) to compute \(\Lambda^0_{\text{HCOOH}}\). 

Step 1: Apply Kohlrausch’s Law to Each Electrolyte

For each electrolyte, we write the molar conductivity as the sum of the ionic conductivities:

1. Na$_2$SO$_4$: Dissociates as \(\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}\): \[ \Lambda^0_{\text{Na}_2\text{SO}_4} = 2\lambda^0_{\text{Na}^+} + \lambda^0_{\text{SO}_4^{2-}} = 260 \]
2. K$_2$SO$_4$: Dissociates as \(\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}\): \[ \Lambda^0_{\text{K}_2\text{SO}_4} = 2\lambda^0_{\text{K}^+} + \lambda^0_{\text{SO}_4^{2-}} = 308 \]
3. KCl: Dissociates as \(\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-\): \[ \Lambda^0_{\text{KCl}} = \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} = 150 \]
4. HCl: Dissociates as \(\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-\): \[ \Lambda^0_{\text{HCl}} = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} = 426 \]
5. HCOONa: Dissociates as \(\text{HCOONa} \rightarrow \text{Na}^+ + \text{HCOO}^-\): \[ \Lambda^0_{\text{HCOONa}} = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \]

Step 2: Calculate Ionic Conductivities
 We need \(\lambda^0_{\text{H}^+}\) and \(\lambda^0_{\text{HCOO}^-}\). Let’s solve for these using the given equations.
Find \(\lambda^0_{\text{HCOO}^-}\):} From the equation for HCOONa: \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \tag{5} \] We need \(\lambda^0_{\text{Na}^+}\). Use the equations for Na$_2$SO$_4$ and K$_2$SO$_4$ to find \(\lambda^0_{\text{Na}^+}\): \[ 2\lambda^0_{\text{Na}^+} + \lambda^0_{\text{SO}_4^{2-}} = 260 \tag{1} \] \[ 2\lambda^0_{\text{K}^+} + \lambda^0_{\text{SO}_4^{2-}} = 308 \tag{2} \] Subtract (1) from (2) to eliminate \(\lambda^0_{\text{SO}_4^{2-}}\): \[ (2\lambda^0_{\text{K}^+} + \lambda^0_{\text{SO}_4^{2-}}) - (2\lambda^0_{\text{Na}^+} + \lambda^0_{\text{SO}_4^{2-}}) = 308 - 260 \] \[ 2\lambda^0_{\text{K}^+} - 2\lambda^0_{\text{Na}^+} = 48 \] \[ \lambda^0_{\text{K}^+} - \lambda^0_{\text{Na}^+} = 24 \tag{6} \] Now, use the equation for KCl: \[ \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} = 150 \tag{3} \] We need \(\lambda^0_{\text{K}^+}\). Let’s find \(\lambda^0_{\text{Cl}^-}\) using HCl: \[ \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} = 426 \tag{4} \] We don’t have \(\lambda^0_{\text{H}^+}\) yet, so let’s try to find \(\lambda^0_{\text{Na}^+}\) and \(\lambda^0_{\text{K}^+}\) first. From (6): \[ \lambda^0_{\text{K}^+} = \lambda^0_{\text{Na}^+} + 24 \] Substitute into (3): \[ (\lambda^0_{\text{Na}^+} + 24) + \lambda^0_{\text{Cl}^-} = 150 \] \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-} = 126 \tag{7} \] Now we have two equations involving \(\lambda^0_{\text{Na}^+}\): \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \tag{5} \] \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-} = 126 \tag{7} \] To find \(\lambda^0_{\text{HCOO}^-}\), we need \(\lambda^0_{\text{Na}^+}\). Let’s find \(\lambda^0_{\text{Cl}^-}\) and \(\lambda^0_{\text{K}^+}\) to proceed. Substitute \(\lambda^0_{\text{K}^+} = \lambda^0_{\text{Na}^+} + 24\) into the equation for K$_2$SO$_4$ and combine with Na$_2$SO$_4$. Instead, let’s try to find \(\lambda^0_{\text{Cl}^-}\) and \(\lambda^0_{\text{H}^+}\) directly. 
Find \(\lambda^0_{\text{H}^+}\) and \(\lambda^0_{\text{Cl}^-}\):
From HCl: \[ \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} = 426 \tag{4} \] From KCl: \[ \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} = 150 \tag{3} \] We need \(\lambda^0_{\text{K}^+}\). Use (6) and solve later. Let’s try to find \(\lambda^0_{\text{Na}^+}\) and \(\lambda^0_{\text{HCOO}^-}\) first. Subtract (7) from (5): \[ (\lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-}) - (\lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-}) = 105 - 126 \] \[ \lambda^0_{\text{HCOO}^-} - \lambda^0_{\text{Cl}^-} = -21 \] \[ \lambda^0_{\text{Cl}^-} = \lambda^0_{\text{HCOO}^-} + 21 \tag{8} \] Substitute into (7): \[ \lambda^0_{\text{Na}^+} + (\lambda^0_{\text{HCOO}^-} + 21) = 126 \] \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \tag{9} \] This is consistent with (5). Let’s find \(\lambda^0_{\text{K}^+}\) using (3) and (8): \[ \lambda^0_{\text{K}^+} + (\lambda^0_{\text{HCOO}^-} + 21) = 150 \] \[ \lambda^0_{\text{K}^+} + \lambda^0_{\text{HCOO}^-} = 129 \tag{10} \] Now use (6): \[ \lambda^0_{\text{K}^+} = \lambda^0_{\text{Na}^+} + 24 \] Substitute into (10): \[ (\lambda^0_{\text{Na}^+} + 24) + \lambda^0_{\text{HCOO}^-} = 129 \] \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \tag{11} \] This matches (5), confirming consistency. Let’s find \(\lambda^0_{\text{Na}^+}\) using Na$_2$SO$_4$ and K$_2$SO$_4$. Substitute \(\lambda^0_{\text{K}^+} = \lambda^0_{\text{Na}^+} + 24\) into (2): \[ 2(\lambda^0_{\text{Na}^+} + 24) + \lambda^0_{\text{SO}_4^{2-}} = 308 \] \[ 2\lambda^0_{\text{Na}^+} + 48 + \lambda^0_{\text{SO}_4^{2-}} = 308 \] \[ 2\lambda^0_{\text{Na}^+} + \lambda^0_{\text{SO}_4^{2-}} = 260 \tag{12} \] Compare with (1): \[ 2\lambda^0_{\text{Na}^+} + \lambda^0_{\text{SO}_4^{2-}} = 260 \] This suggests we need to solve for \(\lambda^0_{\text{Na}^+}\) and \(\lambda^0_{\text{HCOO}^-}\) directly. Let’s try substituting known values. Assume: \[ \lambda^0_{\text{Na}^+} + \lambda^0_{\text{HCOO}^-} = 105 \tag{5} \] Use (8) in (4): \[ \lambda^0_{\text{H}^+} + (\lambda^0_{\text{HCOO}^-} + 21) = 426 \] \[ \lambda^0_{\text{H}^+} + \lambda^0_{\text{HCOO}^-} = 405 \tag{13} \] Now we have: \[ \Lambda^0_{\text{HCOOH}} = \lambda^0_{\text{H}^+} + \lambda^0_{\text{HCOO}^-} = 405 \] Let’s verify by calculating \(\lambda^0_{\text{Na}^+}\), \(\lambda^0_{\text{K}^+}\), and \(\lambda^0_{\text{Cl}^-}\). 
Step 3: Verify Calculations
Assume \(\lambda^0_{\text{HCOO}^-} = x\), then from (5): \[ \lambda^0_{\text{Na}^+} = 105 - x \] From (8): \[ \lambda^0_{\text{Cl}^-} = x + 21 \] From (6) and (3): \[ \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} = 150 \] \[ \lambda^0_{\text{K}^+} = \lambda^0_{\text{Na}^+} + 24 = (105 - x) + 24 = 129 - x \] Substitute into (3): \[ (129 - x) + (x + 21) = 150 \] \[ 129 + 21 = 150 \] This confirms consistency. Now use Na$_2$SO$_4$: \[ 2(105 - x) + \lambda^0_{\text{SO}_4^{2-}} = 260 \] \[ 210 - 2x + \lambda^0_{\text{SO}_4^{2-}} = 260 \] \[ \lambda^0_{\text{SO}_4^{2-}} = 50 + 2x \tag{14} \] For K$_2$SO$_4$: \[ 2(129 - x) + (50 + 2x) = 308 \] \[ 258 - 2x + 50 + 2x = 308 \] \[ 308 = 308 \] This is consistent. Now find \(\lambda^0_{\text{H}^+}\): \[ \lambda^0_{\text{H}^+} + (x + 21) = 426 \] \[ \lambda^0_{\text{H}^+} = 405 - x \] \[ \lambda^0_{\text{H}^+} + \lambda^0_{\text{HCOO}^-} = (405 - x) + x = 405 \] This confirms: \[ \Lambda^0_{\text{HCOOH}} = 405 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1} \] Step 4: Final Answer The molar conductivity at infinite dilution for formic acid is: \[ \boxed{405 \, \text{S}\cdot\text{cm}^2\cdot\text{mol}^{-1}} \]