The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a co

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

The factor of 1000 is a unit conversion factor. It converts the concentration from moles per liter (M) to moles per cubic centimeter, ensuring the final units of molar conductivity are in $\text{S cm}^2\text{ mol}^{-1}$.

Q: Can we use $[H^+] = \sqrt{K_a C}$ here?

While $[H^+] = \sqrt{K_a C}$ is valid for weak acids, we don't have the dissociation constant $K_a$ given in the problem. Instead, we use the fundamental definition $[H^+] = C\alpha$ which directly links to the degree of dissociation we need.

Q: Why is the degree of dissociation $\alpha$ equal to $\Lambda_m / \Lambda_m^0$?

According to Arrhenius theory of electrolytic dissociation, the increase in molar conductivity with dilution for weak electrolytes is entirely due to an increase in the degree of dissociation. At infinite dilution, dissociation is 100% ($\alpha=1$), so the ratio of conductivity at any concentration to infinite dilution gives the fraction dissociated.

Q: What does platinized Pt electrodes mean and why are they used?

Platinized Pt electrodes are coated with finely divided platinum black. This vastly increases their surface area, which reduces polarization effects during alternating current measurements, ensuring accurate conductance readings.

Q: Why did we leave $\alpha$ as a fraction instead of calculating the decimal?

Leaving intermediate values as fractions prevents rounding errors and often leads to beautiful cancellations in the final step, just like how $0.0015$ perfectly cancelled out when we calculated $\Lambda_m^0$.

Question

The conductance of a $0.0015	ext{ M}$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is $120	ext{ cm}$ with an area of cross section of $1	ext{ cm}^2$. The conductance of this solution was found to be $5 	imes 10^{-7}	ext{ S}$. The pH of the solution is $4$. The value of limiting molar conductivity ($\Lambda_m^0$) of this weak monobasic acid in aqueous solution is $Z 	imes 10^2	ext{ S cm}^2	ext{ mol}^{-1}$. The value of $Z$ is.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of multiple electrochemical concepts playing together. We are given a conductivity cell filled with a weak monobasic acid, and we need to find its limiting molar conductivity, $\Lambda_m^0$. Let's break down the journey step-by-step.

### Decoding the Conductivity Cell

Imagine the physical setup: a conductivity cell with two platinized platinum electrodes. The distance between them, $l$, is $120	ext{ cm}$, and their cross-sectional area, $A$, is $1	ext{ cm}^2$. The conductance, $G$, is measured as $5 	imes 10^{-7}	ext{ S}$.

First, we need to find the specific conductivity, $\kappa$. The formula connects conductance to the cell geometry:

$$\kappa = G 	imes \frac{l}{A}$$

Plugging in our values, we get:

$$\kappa = 5 	imes 10^{-7} 	imes \frac{120}{1} = 6 	imes 10^{-5}	ext{ S cm}^{-1}$$

This $\kappa$ represents the conducting power of all the ions present in $1	ext{ cm}^3$ of the solution.

### Bridging Conductance to Molar Conductivity

Now that we have $\kappa$, we can calculate the molar conductivity, $\Lambda_m$, at the given concentration $C = 0.0015	ext{ M}$. Molar conductivity tells us the conducting power of all the ions produced by dissolving one mole of the electrolyte in the solution.

The formula is:

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

Substituting our values:

$$\Lambda_m = \frac{6 	imes 10^{-5} 	imes 1000}{0.0015} = \frac{6 	imes 10^{-2}}{1.5 	imes 10^{-3}} = 40	ext{ S cm}^2	ext{ mol}^{-1}$$

### The pH Connection: Unlocking Dissociation

The problem also gives us the pH of the solution, which is $4$. This is a crucial piece of the puzzle because it directly gives us the hydrogen ion concentration:

$$[H^+] = 10^{-	ext{pH}} = 10^{-4}	ext{ M}$$

For a weak monobasic acid (let's call it HX), the dissociation can be represented as $	ext{HX} ightleftharpoons 	ext{H}^+ + 	ext{X}^-$. If the initial concentration is $C$ and the degree of dissociation is $\alpha$, the equilibrium concentration of $	ext{H}^+$ is $C\alpha$. Therefore:

$$[H^+] = C\alpha$$

We can rearrange this to find $\alpha$:

$$\alpha = \frac{[H^+]}{C} = \frac{10^{-4}}{0.0015}$$

We will leave $\alpha$ in this fractional form to avoid messy decimals and rounding errors in our final calculation.

### The Grand Finale: Arrhenius Meets Kohlrausch

According to Arrhenius theory, the degree of dissociation $\alpha$ is the ratio of molar conductivity at a given concentration to the limiting molar conductivity at infinite dilution:

$$\alpha = \frac{\Lambda_m}{\Lambda_m^0}$$

Rearranging this to solve for our target, $\Lambda_m^0$:

$$\Lambda_m^0 = \frac{\Lambda_m}{\alpha}$$

Now, we substitute the values we've painstakingly calculated:

$$\Lambda_m^0 = \frac{40}{\left(\frac{10^{-4}}{0.0015}ight)}$$

Flipping the denominator, we get:

$$\Lambda_m^0 = \frac{40 	imes 0.0015}{10^{-4}} = \frac{0.06}{10^{-4}} = 600	ext{ S cm}^2	ext{ mol}^{-1}$$

The problem states that $\Lambda_m^0 = Z 	imes 10^2	ext{ S cm}^2	ext{ mol}^{-1}$. Comparing this with our result of $600$, which is $6 	imes 10^2$, we can clearly see that:

**$Z = 6$**

This problem elegantly weaves together the physical geometry of a conductivity cell, the chemical equilibrium of weak acids, and the fundamental laws of electrochemistry.

Enter Numerical Value:

Visualized Solution

$Conductivity Cell Setup$

$Specific Conductivity (κ)$

$Calculating κ$

$Molar Conductivity (Λ_{m})$

$Calculating Λ_{m}$

$Degree of Dissociation (α)$

$Calculating α$

$Limiting Molar Conductivity (Λ_{m}^{0})$

$Substituting Values$

$Calculating Λ_{m}^{0}$

$Finding Z$

$The Way Forward$

The Sigma Insight: Electrolytic Conduction

Decoding the Conductivity Cell

Bridging Conductance to Molar Conductivity

The pH Connection

The Grand Finale

Similar Questions

Comprehension Passage

The value of $α$ is ______.

The value of $y$ is ______.

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$Λ_{m}^{\circ}$ 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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A $5.0 m mol dm^{- 3}$ aqueous solution of KCl has a conductance of $0.55 mS$ when measured in a cell constant $1.3 cm^{- 1}$ . The molar conductivity of this solution is .......... $mS m^{2} mol^{- 1}$ . (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Specific Conductivity (κ)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Conductivity (Λm​)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Degree of Dissociation (α)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Molar Conductivity (Λm0​)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Λm0​

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

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