Ionic conductances 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 infinite dilution is:

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 \( H_2SO_4 \)
The first step is to recognize that sulfuric acid (\( H_2SO_4 \)) dissociates in water to form two hydrogen ions (\( H^+ \)) and one sulfate ion (\( SO_4^{2-} \)). The dissociation can be represented as:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]

### Step 2: Write the formula for equivalent conductance
The equivalent conductance (\( \Lambda_{eq} \)) at infinite dilution can be calculated using the formula:
\[ \Lambda_{eq} = n_1 \cdot \lambda_{H^+} + n_2 \cdot \lambda_{SO_4^{2-}} \]
where:
- \( n_1 \) and \( n_2 \) are the stoichiometric coefficients of the ions in the dissociation (which are 2 for \( H^+ \) and 1 for \( SO_4^{2-} \)).
- \( \lambda_{H^+} \) is the ionic conductance of \( H^+ \) (given as \( x \)).
- \( \lambda_{SO_4^{2-}} \) is the ionic conductance of \( SO_4^{2-} \) (given as \( y \)).

### Step 3: Substitute the values into the formula
Substituting the values into the equivalent conductance formula:
\[ \Lambda_{eq} = 2 \cdot x + 1 \cdot y \]
This simplifies to:
\[ \Lambda_{eq} = 2x + y \]

### Step 4: Divide by the total number of equivalents
Since \( H_2SO_4 \) can donate 2 equivalents of \( H^+ \) ions, we divide the total conductance by the number of equivalents (which is 2):
\[ \Lambda_{eq} = \frac{2x + y}{2} \]

### Final Answer
Thus, the equivalent conductance of \( H_2SO_4 \) at infinite dilution is:
\[ \Lambda_{eq} = \frac{2x + y}{2} \, \text{S cm}^{-2} \text{equiv}^{-1} \]

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