The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move

Q: Why can't we measure the limiting molar conductivity of acetic acid directly?

Because it's a weak electrolyte, its degree of dissociation doesn't reach 100% even at very high dilutions, making extrapolation to zero concentration on a graph highly inaccurate.

Q: What exactly is Kohlrausch's Law?

It states that at infinite dilution, each ion migrates independently of its co-ion and contributes a definite, constant value to the total molar conductivity of the electrolyte.

Q: Why do we add $\text{CH}_3\text{COONa}$ and $\text{HCl}$ together?

We add them because they are strong electrolytes that contain the specific ions we need: $\text{CH}_3\text{COO}^-$ from sodium acetate and $\text{H}^+$ from hydrochloric acid.

Q: What happens to the $\text{Na}^+$ and $\text{Cl}^-$ ions?

They act as 'spectator' ions in this mathematical manipulation. Because we don't want them in our final acetic acid equation, we must subtract their combined contribution, which is why we need the conductivity of $\text{NaCl}$.

Q: Are equivalent and molar conductivities the same here?

Yes! For uni-univalent electrolytes like $\text{NaCl}$, $\text{HCl}$, and $\text{CH}_3\text{COONa}$, the n-factor (total positive or negative charge) is 1, so their equivalent conductance is numerically equal to their molar conductance.

Question

The equivalent conductances of two strong electrolytes at infinite dilution in $	ext{H}_2	ext{O}$ (where ions move freely through a solution) at $25^\circ	ext{C}$ are given below
$\Lambda^\circ_{	ext{CH}_3	ext{COONa}} = 91.0 	ext{ S cm}^2/	ext{equiv}$
$\Lambda^\circ_{	ext{HCl}} = 426.2 	ext{ S cm}^2/	ext{equiv}$
What additional information/quantity one needs to calculate $\Lambda^\circ$ of an aqueous solution of acetic acid ?

Options:

(A) $\Lambda^\circ$ of $	ext{NaCl}$

(B) $\Lambda^\circ$ of $	ext{CH}_3	ext{COOK}$

(C) The limiting equivalent conductance of $	ext{H}^+$ ($\lambda^\circ_{	ext{H}^+}$)

(D) $\Lambda^\circ$ of chloroacetic acid ($	ext{ClCH}_2	ext{COOH}$)

Sigma Percentile · Accepted Answer

### The Challenge of Weak Electrolytes

Imagine trying to measure the top speed of a car that never fully accelerates. That is exactly the problem we face with weak electrolytes like acetic acid ($	ext{CH}_3	ext{COOH}$). Because it never fully dissociates into ions, even at very high dilutions, we cannot directly measure its limiting molar conductivity ($\Lambda^\circ$) by simply extrapolating a graph to zero concentration.

So, how do we find it? We use a clever mathematical trick provided by **Kohlrausch's Law of Independent Migration of Ions**.

### Kohlrausch's Law to the Rescue

Kohlrausch's Law states that at infinite dilution, every ion migrates independently of its co-ion. This means the total limiting molar conductivity of an electrolyte is simply the sum of the individual limiting ionic conductivities:

$\Lambda^\circ_{	ext{CH}_3	ext{COOH}} = \lambda^\circ_{	ext{CH}_3	ext{COO}^-} + \lambda^\circ_{	ext{H}^+}$

To find this value, we can use strong electrolytes—which *do* fully dissociate and whose $\Lambda^\circ$ values are easy to measure. The problem gives us two such strong electrolytes:

1. Sodium Acetate: $\Lambda^\circ_{	ext{CH}_3	ext{COONa}} = \lambda^\circ_{	ext{CH}_3	ext{COO}^-} + \lambda^\circ_{	ext{Na}^+}$
2. Hydrochloric Acid: $\Lambda^\circ_{	ext{HCl}} = \lambda^\circ_{	ext{H}^+} + \lambda^\circ_{	ext{Cl}^-}$

### The Mathematical Puzzle

If we add the conductances of these two strong electrolytes together, we get the ions we want (acetate and hydrogen), but we also get some extra "spectator" ions:

$\Lambda^\circ_{	ext{CH}_3	ext{COONa}} + \Lambda^\circ_{	ext{HCl}} = (\lambda^\circ_{	ext{CH}_3	ext{COO}^-} + \lambda^\circ_{	ext{H}^+}) + (\lambda^\circ_{	ext{Na}^+} + \lambda^\circ_{	ext{Cl}^-})$

Notice the extra term at the end? That is exactly the limiting molar conductivity of sodium chloride ($	ext{NaCl}$). To isolate the value for acetic acid, we must subtract the contribution of these unwanted $	ext{Na}^+$ and $	ext{Cl}^-$ ions.

$\Lambda^\circ_{	ext{CH}_3	ext{COOH}} = \Lambda^\circ_{	ext{CH}_3	ext{COONa}} + \Lambda^\circ_{	ext{HCl}} - \Lambda^\circ_{	ext{NaCl}}$

Therefore, the additional piece of information we desperately need to complete this calculation is the limiting equivalent conductance of **$	ext{NaCl}$**.

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Visualized Solution

$Kohlrausch’s Law$

$Target Equation$

$Given Electrolytes$

$Mathematical Manipulation$

$Finding the Missing Piece$

The Sigma Insight: Electrolytic Conduction

The Challenge of Weak Electrolytes

Kohlrausch's Law to the Rescue

The Mathematical Puzzle

Similar Questions

Calculate $Λ_{HOAc}^{\infty}$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_{2} O$ at $2 5^{\circ} C$ . $Electrolyte Λ_{m}^{\infty} (S cm^{2} mol^{- 1}) KCl 149.9 KNO_{3} 145.0 HCl 426.2 NaOAc 91.0 NaCl 126.5$

The molar conductivities $Λ_{NaOAc}^{\circ}$ and $Λ_{HCl}^{\circ}$ at infinite dilution in water at $2 5^{\circ} C$ are $91.0$ and $426.2 S cm^{2} / mol$ , respectively. To calculate $Λ_{HOAc}^{\circ}$ , the additional value required is

$Λ_{m}^{\circ}$ 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

The equivalent conductance of NaCl at concentration $C$ and at infinite dilution are $λ_{c}$ and $λ_{\infty}$ , respectively. The correct relationship between $λ_{c}$ and $λ_{\infty}$ is given as

Comprehension Passage

The value of $α$ is ______.

The value of $y$ is ______.

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are $280$ , $860$ and $426 S cm^{2} mol^{- 1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is ... $S cm^{2} mol^{- 1}$ (Round off to the nearest Integer).

The conductivity of a weak acid HA of concentration $0.001 mol L^{- 1}$ is $2.0 \times 1 0^{- 5} S cm^{- 1}$ . If $Λ_{m}^{\circ} (HA) = 190 S cm^{2} mol^{- 1}$ , the ionisation constant ( $K_{a}$ ) of HA is equal to ............ $\times 1 0^{- 6}$ . (Round off to the nearest integer)

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kohlrausch’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Target Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Electrolytes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mathematical Manipulation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Missing Piece