Ionic conductance of `H^(+) and SO_(4)""( 2–)` at infinite dilution are x and y S `"cm"^(2)" equiv "^(–1)`. Hence, equivalent conductance of `H_(2) SO_(4)` at inifinite dilution will be:

Options

A. x + y
B. 2(x + y)
C. 2x + y
D. x + 2y

Answer

To find the equivalent conductance of \( H_2SO_4 \) at infinite dilution, we can follow these steps:

### Step 1: Understand the dissociation of \( H_2SO_4 \)
When sulfuric acid (\( H_2SO_4 \)) dissociates in water, it produces:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
This means that for every molecule of \( H_2SO_4 \), we get 2 moles of \( H^+ \) ions and 1 mole of \( SO_4^{2-} \) ions.

### Step 2: Define equivalent conductance
The equivalent conductance (\( \Lambda_{eq} \)) is defined as the conductance of one equivalent of an electrolyte in solution. It can be calculated using the formula:
\[ \Lambda_{eq} = \frac{\text{ionic conductance of } H^+}{n_{H^+}} \cdot \text{coefficient of } H^+ + \frac{\text{ionic conductance of } SO_4^{2-}}{n_{SO_4^{2-}}} \cdot \text{coefficient of } SO_4^{2-} \]

### Step 3: Identify the n-factors
- For \( H^+ \), the n-factor is 1 (since it has a +1 charge).
- For \( SO_4^{2-} \), the n-factor is 2 (since it has a -2 charge).

### Step 4: Substitute the values
Let the ionic conductance of \( H^+ \) be \( x \) S cm\(^2\) equiv\(^{-1}\) and that of \( SO_4^{2-} \) be \( y \) S cm\(^2\) equiv\(^{-1}\).

Now substituting these into the equivalent conductance formula:
\[ \Lambda_{eq} = \frac{x}{1} \cdot 2 + \frac{y}{2} \cdot 1 \]
This simplifies to:
\[ \Lambda_{eq} = 2x + \frac{y}{2} \]

### Step 5: Final expression
Thus, the equivalent conductance of \( H_2SO_4 \) at infinite dilution is:
\[ \Lambda_{eq} = 2x + \frac{y}{2} \]