At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentr

Q: Why do we need to extrapolate the graph for AgCl?

We actually don't! AgCl is a weak/sparingly soluble electrolyte, so its graph is a steep curve that never touches the y-axis. That's exactly why we use Kohlrausch's law of independent migration of ions to calculate it indirectly using strong electrolytes.

Q: Why is the constant 'b' different for AgNO3 compared to NaCl and NaNO3?

The constant 'b' depends on the nature of the electrolyte (its stoichiometry, like 1:1, 2:1) and the specific interactions between its ions. While NaCl, NaNO3, and AgNO3 are all 1:1 electrolytes, their specific ionic sizes and mobilities cause slight variations in 'b'.

Q: Why do we assume $\Lambda_m = \Lambda_m^\circ$ for the saturated AgCl solution?

Because AgCl is sparingly soluble, its saturated solution is extremely dilute (concentration approaches zero). At such infinite dilution, the ions are so far apart that they don't interact, meaning its molar conductivity is practically equal to its limiting molar conductivity.

Q: What is the significance of the $1000$ in the formula $\Lambda_m = \frac{\kappa \times 1000}{M}$?

It's a unit conversion factor! Conductivity $\kappa$ is usually given in $\text{S cm}^{-1}$, while molarity $M$ is in $\text{mol L}^{-1}$ (or $\text{mol dm}^{-3}$). Multiplying by $1000$ converts the volume from liters to cubic centimeters, ensuring the units match up perfectly to give $\Lambda_m$ in $\text{S cm}^2 \text{ mol}^{-1}$.

Q: Can I just subtract the conductivities directly without finding $\Lambda_m^\circ$ first?

No, that's a classic trap! Kohlrausch's law of independent migration strictly applies only at infinite dilution ($\Lambda_m^\circ$). At any finite concentration, ion-ion interactions interfere, and the conductivities are not perfectly additive.

Question

At $300 	ext{ K}$, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below :
\begin{array}{|c|c|c|}
\hline
	ext{Salt} & 	ext{Concentration (M)} & 	ext{Molar conductivity } (	ext{S cm}^2 	ext{ mol}^{-1}) \
\hline
	ext{NaNO}_3 & 0.01 & 111 \
 & 0.04 & 101 \
\hline
	ext{NaCl} & 0.01 & 117 \
 & 0.04 & 107 \
\hline
	ext{AgNO}_3 & 0.01 & 125 \
 & 0.04 & 116 \
\hline
\end{array}

The conductivity of a saturated aqueous solution of $	ext{AgCl}$ is $1.40 	imes 10^{-6} 	ext{ S cm}^{-1}$ at $300 	ext{ K}$. If the solubility of $	ext{AgCl}$ in water at $300 	ext{ K}$ is $X 	ext{ mol L}^{-1}$, then $\log_{10}(X^{-1})$ is

(Assume that $	ext{AgCl}$ dissolved in water ionizes completely and that the molar conductivity of saturated $	ext{AgCl}$ solution is equal to its limiting molar conductivity.)

Options:

(A) 3

(B) 4

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### The Challenge of Sparingly Soluble Salts

Imagine you are tasked with finding the solubility of silver chloride ($	ext{AgCl}$). It's a notoriously stubborn salt that barely dissolves in water. Because it's so sparingly soluble, directly measuring its concentration is incredibly difficult.

However, electrochemistry gives us a brilliant backdoor. If we can determine its **limiting molar conductivity** ($\Lambda_m^\circ$), we can unlock its solubility using the relationship between conductivity and concentration. But there's a catch: because $	ext{AgCl}$ is a weak electrolyte in practice, we cannot simply plot its conductivity against concentration and extrapolate to zero. The curve would shoot up infinitely!

### The Magic of Kohlrausch's Law

To bypass this, we rely on **Kohlrausch's Law of Independent Migration of Ions**. This law states that at infinite dilution, every ion migrates independently of its counter-ion. Therefore, we can construct the limiting molar conductivity of $	ext{AgCl}$ using a clever combination of strong electrolytes:

$$\Lambda_{m(	ext{AgCl})}^\circ = \Lambda_{m(	ext{AgNO}_3)}^\circ + \Lambda_{m(	ext{NaCl})}^\circ - \Lambda_{m(	ext{NaNO}_3)}^\circ$$

Notice how the $	ext{Na}^+$ and $	ext{NO}_3^-$ ions mathematically cancel out, leaving us perfectly with $	ext{Ag}^+$ and $	ext{Cl}^-$! Our new mission is simply to find the $\Lambda_m^\circ$ values for these three strong electrolytes.

### Extrapolating the Strong Electrolytes

For strong electrolytes, Kohlrausch gave us another powerful tool—a linear equation relating molar conductivity to the square root of concentration:

$$\Lambda_m = \Lambda_m^\circ - b\sqrt{c}$$

We are given data at two concentrations ($c = 0.01 	ext{ M}$ and $c = 0.04 	ext{ M}$) for each salt. Let's set up the equations.

**For $	ext{NaNO}_3$:**
Taking the square roots of the concentrations gives $0.1$ and $0.2$. 
$$111 = \Lambda_{m(	ext{NaNO}_3)}^\circ - 0.1b$$
$$101 = \Lambda_{m(	ext{NaNO}_3)}^\circ - 0.2b$$
Subtracting the second equation from the first yields $10 = 0.1b$, so $b = 100$. Substituting this back gives $\Lambda_{m(	ext{NaNO}_3)}^\circ = 121 	ext{ S cm}^2 	ext{ mol}^{-1}$.

**For $	ext{NaCl}$:**
$$117 = \Lambda_{m(	ext{NaCl})}^\circ - 0.1b$$
$$107 = \Lambda_{m(	ext{NaCl})}^\circ - 0.2b$$
Solving this similarly gives $\Lambda_{m(	ext{NaCl})}^\circ = 127 	ext{ S cm}^2 	ext{ mol}^{-1}$.

**For $	ext{AgNO}_3$:**
$$125 = \Lambda_{m(	ext{AgNO}_3)}^\circ - 0.1b$$
$$116 = \Lambda_{m(	ext{AgNO}_3)}^\circ - 0.2b$$
Solving this gives $\Lambda_{m(	ext{AgNO}_3)}^\circ = 134 	ext{ S cm}^2 	ext{ mol}^{-1}$.

### Assembling the Puzzle

Now, we plug these extrapolated values back into our master equation:

$$\Lambda_{m(	ext{AgCl})}^\circ = 134 + 127 - 121 = 140 	ext{ S cm}^2 	ext{ mol}^{-1}$$

We have successfully found the limiting molar conductivity of our sparingly soluble salt!

### The Final Calculation: Unlocking Solubility

For a saturated solution of a sparingly soluble salt, the concentration is so incredibly low that we can safely assume its molar conductivity is practically equal to its limiting molar conductivity ($\Lambda_m \approx \Lambda_m^\circ$). The formula connecting them is:

$$\Lambda_m^\circ = \frac{\kappa 	imes 1000}{S}$$

Where $\kappa$ is the specific conductivity and $S$ is the solubility in $	ext{mol L}^{-1}$. Let's substitute our known values:

$$140 = \frac{1.40 	imes 10^{-6} 	imes 1000}{S}$$

Rearranging to solve for $S$:

$$S = \frac{1.40 	imes 10^{-3}}{140} = 10^{-5} 	ext{ M}$$

The problem defines this solubility as $X$. Therefore, $X = 10^{-5}$.

The final question asks for $\log_{10}(X^{-1})$.

$$X^{-1} = (10^{-5})^{-1} = 10^5$$

$$\log_{10}(10^5) = 5$$

And there we have it! A beautiful journey from raw conductivity data, through graphical extrapolation and ionic migration, all the way to the precise solubility of a stubborn salt.

Select Answer:

Visualized Solution

$Visualizing the Goal$

$Kohlrausch’s Law of Independent Migration$

$Kohlrausch’s Equation for Strong Electrolytes$

$Calculating Λ_{m}^{\circ} for NaNO_{3}$

$Calculating Λ_{m}^{\circ} for NaCl$

$Calculating Λ_{m}^{\circ} for AgNO_{3}$

$Finding Λ_{m}^{\circ} for AgCl$

$Relating Conductivity to Solubility$

$Substituting the Values$

$Calculating Solubility S$

$Final Calculation$

The Sigma Insight: Electrolytic Conduction

The Challenge of Sparingly Soluble Salts

The Magic of Kohlrausch's Law

Extrapolating the Strong Electrolytes

Assembling the Puzzle

The Final Calculation

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Select Answer:

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Λm∘​ for NaNO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Λm∘​ for NaCl

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Λm∘​ for AgNO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Λm∘​ for AgCl

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating Conductivity to Solubility

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

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

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