Animated Solution for Chemistry - Electrochemistry: A KCl solution of conductivity $0.14{\text{ S m}}^{-1}$ shows a resistance of $4.19\Omega$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to $1.03\Omega$. The conductivity of the HCl solution is ......... $\times 10^{-2}{\text{ S m}}^{-1}$ (Round off to the nearest integer)

Mastering Conductivity

Imagine you are standing in a laboratory, holding a small, delicate piece of glassware. Inside this glass vessel are two metal plates, perfectly parallel to each other. This is a conductivity cell, a simple yet profound tool used to measure how easily electricity flows through a liquid.

In this problem, we are going to perform a classic thought experiment. We will fill this cell with one solution, take a measurement, and then swap it out for another. The secret to unlocking this question lies not in what changes, but in what stays exactly the same. Let's dive into the fascinating world of electrolytic conduction!

Before we crunch any numbers, we need to understand the physical reality of our tools. A conductivity cell has two electrodes. The distance between these electrodes is $l$, and their cross-sectional area is $A$.

No matter what liquid you pour into this cell—whether it's a potassium chloride (KCl) solution, hydrochloric acid (HCl), or just pure water—the physical dimensions of the glass and metal do not change. The ratio of the distance to the area, $\frac{l}{A}$, is a fixed geometric property of the cell. We call this the cell constant, denoted by $G^{\ast }$.

To connect our physical cell to the measurements we take, we need a bridge. That bridge is the master equation of conductivity:

Let's break this down. $\kappa$ (kappa) is the conductivity of the solution. It tells us how well the ions in the liquid can carry an electric current. $R$ is the resistance, which is what our meter actually measures. And $G^{\ast }$ is our trusty cell constant.

Notice the relationship here. If we rearrange the equation, we get:

This tells us that for any given cell, the product of the solution's conductivity and its measured resistance will always equal the cell constant.

Now, let's look at the narrative of our problem. First, we fill the cell with a KCl solution. We are given its conductivity, ${\kappa }_{\text{KCl}}=0.14{\text{ S m}}^{-1}$, and we measure its resistance, $R_{\text{KCl}}=4.19\Omega$.

Next, we empty the cell, clean it, and fill it with an HCl solution. We measure a new resistance, $R_{\text{HCl}}=1.03\Omega$.

Why did the resistance drop so dramatically? Because HCl is a strong acid, and its $H^{+}$ ions are incredibly fast and mobile compared to the heavier $K^{+}$ ions. The liquid is now a much better conductor!

But remember our secret: the cell itself hasn't changed. The distance between the plates $l$ and their area $A$ are identical. Therefore, the cell constant $G^{\ast }$ for the KCl measurement is exactly the same as the cell constant for the HCl measurement.

Since $G^{\ast }$ is constant, we can set up a beautiful equivalence. The product of conductivity and resistance for KCl must equal the product of conductivity and resistance for HCl:

This is the pivotal moment of the problem. We have transformed a physical action (swapping liquids) into a rigorous mathematical relationship.

Our goal is to find the conductivity of the HCl solution, ${\kappa }_{\text{HCl}}$. Let's rearrange our equivalence to isolate our unknown variable:

Now, we carefully substitute the values given in the problem:

Take a breath here. Don't rush the arithmetic. Multiplying the numerator gives us $0.5866$. Dividing that by $1.03$ yields:

We have our answer, but we must respect the format requested by the examiners. The question asks for the value in the form of $\cdots \times 10^{-2}{\text{ S m}}^{-1}$.

To convert our decimal into this format, we shift the decimal point two places to the right and multiply by $10^{-2}$:

Finally, we are asked to round off to the nearest integer. Looking at $56.95$, the first decimal digit is $9$, which means we round up.

The integer we need to enter is 57.

By understanding the physical reality of the cell constant, we turned a potentially confusing scenario into a straightforward algebraic equivalence. Always look for the hidden constants in a changing system—they are the anchors that will guide you to the solution!