Λ°_m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm^2 mol^-1, respectively. If the conductivi

Q: Why do we add and subtract the limiting molar conductivities of the strong electrolytes?

We do this to isolate the specific ions of the weak acid. By adding $\text{HCl}$ and $\text{NaA}$, we get $\text{H}^+$, $\text{Cl}^-$, $\text{Na}^+$, and $\text{A}^-$. Subtracting $\text{NaCl}$ removes the unwanted $\text{Na}^+$ and $\text{Cl}^-$ ions, leaving exactly $\Lambda^\circ_m(\text{H}^+) + \Lambda^\circ_m(\text{A}^-)$, which is $\Lambda^\circ_m(\text{HA})$.

Q: Why do we multiply by 1000 in the molar conductivity formula?

The factor of 1000 is a unit conversion factor. Molarity is typically given in moles per liter ($\text{mol L}^{-1}$), but specific conductivity is often in $\text{S cm}^{-1}$. Since $1 \text{ L} = 1000 \text{ cm}^3$, multiplying by 1000 converts the volume unit so that the final molar conductivity is in $\text{S cm}^2 \text{ mol}^{-1}$.

Q: Can the degree of dissociation $\alpha$ ever be greater than 1?

No, $\alpha$ represents a fraction of the total molecules that have dissociated. It ranges from 0 (no dissociation) to 1 (complete dissociation). If you calculate an $\alpha > 1$, you've likely made a calculation error or mixed up your units.

Q: What is the difference between specific conductivity and molar conductivity?

Specific conductivity ($\kappa$) is the conductance of a $1 \text{ cm}^3$ volume of the solution. Molar conductivity ($\Lambda_m$) is the conducting power of all the ions produced by dissolving exactly 1 mole of an electrolyte in the solution. Molar conductivity normalizes the conductivity based on the number of moles present.

Q: Why can't we find $\Lambda^\circ_m$ for HA directly from a graph of $\Lambda_m$ vs $\sqrt{C}$?

For strong electrolytes, the graph of $\Lambda_m$ vs $\sqrt{C}$ is a straight line that can be easily extrapolated to zero concentration (infinite dilution). However, for weak electrolytes like $\text{HA}$, the curve becomes very steep at low concentrations and never touches the y-axis, making direct extrapolation impossible. That's why we must use Kohlrausch's Law.

Question

$\Lambda^{\circ}_{m}$ for NaCl, HCl and NaA are $126.4$, $425.9$ and $100.5 	ext{ S cm}^{2} 	ext{ mol}^{-1}$, respectively. If the conductivity of $0.001 	ext{ M}$ HA is $5 	imes 10^{-5} 	ext{ S cm}^{-1}$, degree of dissociation of HA is

Options:

(A) 0.25

(B) 0.50

(C) 0.75

(D) 0.125

Sigma Percentile · Accepted Answer

### The Quest for the Degree of Dissociation

Imagine a weak acid, $	ext{HA}$, dissolved in water. Unlike strong electrolytes that completely shatter into ions, a weak acid is hesitant. It only partially dissociates. The fraction of the acid that actually breaks apart into ions is called its **degree of dissociation**, denoted by $\alpha$.

To find $\alpha$, we need to compare how well the solution conducts electricity right now versus how well it *could* conduct if it were completely dissociated. Mathematically, this is expressed as:

$$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$$

Here, $\Lambda_m$ is the molar conductivity at the given concentration, and $\Lambda^\circ_m$ is the limiting molar conductivity (the theoretical maximum conductivity at infinite dilution).

### Step 1: Finding the Theoretical Maximum ($\Lambda^\circ_m$)

Because $	ext{HA}$ is a weak acid, we cannot measure its $\Lambda^\circ_m$ directly by extrapolating a graph. Instead, we use **Kohlrausch's Law of Independent Migration of Ions**. This law states that at infinite dilution, each ion contributes a definite amount to the total conductivity, regardless of its partner.

We can cleverly construct the limiting molar conductivity of $	ext{HA}$ using the strong electrolytes provided: $	ext{HCl}$, $	ext{NaA}$, and $	ext{NaCl}$.

$$\Lambda^\circ_m(	ext{HA}) = \Lambda^\circ_m(	ext{H}^+) + \Lambda^\circ_m(	ext{A}^-)$$

By adding the conductivities of $	ext{HCl}$ and $	ext{NaA}$, we get the ions $	ext{H}^+$, $	ext{Cl}^-$, $	ext{Na}^+$, and $	ext{A}^-$. To isolate just $	ext{H}^+$ and $	ext{A}^-$, we subtract the conductivity of $	ext{NaCl}$:

$$\Lambda^\circ_m(	ext{HA}) = \Lambda^\circ_m(	ext{HCl}) + \Lambda^\circ_m(	ext{NaA}) - \Lambda^\circ_m(	ext{NaCl})$$

Substituting the given values:

$$\Lambda^\circ_m(	ext{HA}) = 425.9 + 100.5 - 126.4 = 400 	ext{ S cm}^2 	ext{ mol}^{-1}$$

### Step 2: Finding the Current Conductivity ($\Lambda_m$)

Next, we calculate the actual molar conductivity at the given concentration of $0.001 	ext{ M}$. The formula relates molar conductivity to specific conductivity ($\kappa$) and molarity ($M$):

$$\Lambda_m = \frac{\kappa 	imes 1000}{M}$$

We are given $\kappa = 5 	imes 10^{-5} 	ext{ S cm}^{-1}$ and $M = 0.001 	ext{ M}$ (which is $10^{-3} 	ext{ M}$). Plugging these in:

$$\Lambda_m = \frac{5 	imes 10^{-5} 	imes 1000}{10^{-3}} = \frac{5 	imes 10^{-2}}{10^{-3}} = 50 	ext{ S cm}^2 	ext{ mol}^{-1}$$

### Step 3: The Final Calculation

Now that we have both pieces of the puzzle, finding the degree of dissociation is a simple division:

$$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{50}{400}$$

$$\alpha = \frac{1}{8} = 0.125$$

This means that at this specific concentration, exactly $12.5\%$ of the weak acid molecules have dissociated into ions. The elegance of electrochemistry lies in how macroscopic measurements like conductivity can reveal the microscopic behavior of molecules!

Animated Solution for Chemistry - Electrochemistry: $Λ_{m}^{\circ}$ for NaCl, HCl and NaA are $126.4$ , $425.9$ and $100.5 S cm^{2} mol^{- 1}$ , respectively. If the conductivity of $0.001 M$ HA is $5 \times 1 0^{- 5} S cm^{- 1}$ , degree of dissociation of HA is

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Visualized Solution

\text{Understanding the Goal}

\text{Kohlrausch's Law for } \Lambda^\circ_m(\text{HA})

\text{Calculating } \Lambda^\circ_m(\text{HA})

\text{Formula for Molar Conductivity } \Lambda_m

\text{Calculating } \Lambda_m

\text{Degree of Dissociation } \alpha

\text{What's Next?}

The Sigma Insight: Electrolytic Conduction

The Quest for the Degree of Dissociation

Step 1

Step 2

Step 3

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Comprehension Passage

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