Calculate limiting molar conductivity of HOAc using appropriate molar conductances of the electrolyt

Q: Why can't we measure the limiting molar conductivity of weak electrolytes directly?

Weak electrolytes do not dissociate completely even at very high dilutions. Extrapolating their molar conductivity vs. square root of concentration graph to zero concentration is impossible because the curve becomes asymptotic to the y-axis.

Q: What are spectator ions in this context?

Spectator ions are the ions that are introduced when we use strong electrolytes to find the conductivity of a weak electrolyte, but they are not part of the final weak electrolyte. In this problem, $\text{Na}^+$ and $\text{Cl}^-$ are spectator ions that we must subtract out.

Q: Does Kohlrausch's law apply to strong electrolytes as well?

Yes, absolutely! Kohlrausch's law is a universal principle that applies to all electrolytes at infinite dilution, where interionic interactions are zero and each ion migrates independently.

Q: Why did we ignore KCl and KNO3?

The problem provides extra data to test your understanding. We only need the electrolytes that contain the ions of our target weak electrolyte ($\text{H}^+$ and $\text{OAc}^-$) and the spectator ions that need to be canceled out ($\text{Na}^+$ and $\text{Cl}^-$).

Q: What does the superscript $\infty$ signify?

The superscript $\infty$ (or sometimes a degree symbol $\circ$) denotes 'infinite dilution'. It represents the hypothetical state where the concentration approaches zero, and the ions are infinitely far apart, experiencing no attractive or repulsive forces from each other.

Question

Calculate $\Lambda^\infty_{	ext{HOAc}}$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $	ext{H}_2	ext{O}$ at $25^\circ	ext{C}$.

$\begin{array}{|c|c|c|c|c|c|} \hline 	ext{Electrolyte} & 	ext{KCl} & 	ext{KNO}_3 & 	ext{HCl} & 	ext{NaOAc} & 	ext{NaCl} \ \hline \Lambda^\infty_m (	ext{S cm}^2 	ext{mol}^{-1}) & 149.9 & 145.0 & 426.2 & 91.0 & 126.5 \ \hline \end{array}$

Options:

(A) 217.5

(B) 390.7

(C) 552.7

(D) 517.2

Sigma Percentile · Accepted Answer

### The Mystery of Weak Electrolytes

Imagine you are trying to measure the ultimate potential of a runner, but every time they run, they are tied to a heavy weight. No matter how much you try to lighten the load, they never truly run free. This is exactly the problem chemists faced when trying to measure the **limiting molar conductivity** of weak electrolytes like acetic acid ($	ext{HOAc}$).

Weak electrolytes are stubborn. Even when you dilute them with massive amounts of water, they refuse to dissociate completely into their constituent ions. If you try to plot their molar conductivity against the square root of their concentration, the graph shoots up asymptotically. You can never pinpoint where it crosses the y-axis. The true potential of the ions remains hidden.

### Enter Kohlrausch's Law

But then came Friedrich Kohlrausch, a brilliant physicist who realized something profound. He discovered that at **infinite dilution**, ions are so far apart that they stop caring about each other. They migrate independently. This means the total conductivity of an electrolyte is simply the sum of the individual conductivities of its ions.

$$
\Lambda^\infty_m = \lambda^\infty_+ + \lambda^\infty_-
$$

This is **Kohlrausch's Law of Independent Migration of Ions**. It was a game-changer because it meant we could use strong electrolytes—which dissociate completely and are easy to measure—to uncover the secrets of weak electrolytes.

### The Master Strategy

Our goal is to find the limiting molar conductivity of acetic acid, $\Lambda^\infty_{	ext{HOAc}}$. According to Kohlrausch's law, this is the sum of the conductivities of the hydrogen ion and the acetate ion:

$$
\Lambda^\infty_{	ext{HOAc}} = \lambda^\infty_{	ext{H}^+} + \lambda^\infty_{	ext{OAc}^-}
$$

We don't have these individual values. But look at the data provided! We have strong electrolytes: $	ext{NaOAc}$, $	ext{HCl}$, and $	ext{NaCl}$.

Notice the clever trick we can play here. We can get the acetate ion ($	ext{OAc}^-$) from sodium acetate ($	ext{NaOAc}$), and the hydrogen ion ($	ext{H}^+$) from hydrochloric acid ($	ext{HCl}$).

Let's add their conductivities together:

$$
\Lambda^\infty_{	ext{NaOAc}} + \Lambda^\infty_{	ext{HCl}} = (\lambda^\infty_{	ext{Na}^+} + \lambda^\infty_{	ext{OAc}^-}) + (\lambda^\infty_{	ext{H}^+} + \lambda^\infty_{	ext{Cl}^-})
$$

We have our desired $	ext{H}^+$ and $	ext{OAc}^-$ ions, but we also have unwanted guests: $	ext{Na}^+$ and $	ext{Cl}^-$. These are our **spectator ions**.

### The Final Execution

To get rid of the spectator ions, we simply subtract the conductivity of sodium chloride ($	ext{NaCl}$):

$$
\Lambda^\infty_{	ext{HOAc}} = \Lambda^\infty_{	ext{NaOAc}} + \Lambda^\infty_{	ext{HCl}} - \Lambda^\infty_{	ext{NaCl}}
$$

Now, it's just a matter of plugging in the numbers. We must be careful not to make a silly mistake with the arithmetic.

$$
\Lambda^\infty_{	ext{HOAc}} = 91.0 + 426.2 - 126.5
$$

First, we add the conductivities of our source electrolytes:

$$
91.0 + 426.2 = 517.2 	ext{ S cm}^2 	ext{mol}^{-1}
$$

Finally, we subtract the conductivity of the spectator ions:

$$
517.2 - 126.5 = 390.7 	ext{ S cm}^2 	ext{mol}^{-1}
$$

And there we have it! The limiting molar conductivity of acetic acid is **$390.7 	ext{ S cm}^2 	ext{mol}^{-1}$**. By cleverly combining strong electrolytes, we bypassed the stubborn nature of weak electrolytes and revealed their true potential at infinite dilution.

Select Answer:

Visualized Solution

Kohlrausch's Law

Formulating the Equation

Combining Strong Electrolytes

Substituting Values

Final Calculation

Conclusion

The Sigma Insight: Electrolytic Conduction

The Mystery of Weak Electrolytes

Enter Kohlrausch's Law

The Master Strategy

The Final Execution

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