Ionic conductance of `H^(+) and SO_(4)` are x and yS `cm^(2) mol^(-1)`. Hence equivalent conductivity of `H_(2)SO_(4)` is

Options

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

Answer

To find the equivalent conductivity of \( H_2SO_4 \), we can follow these steps:

### Step 1: Identify the Ionic Conductances
We are given the ionic conductances of \( H^+ \) and \( SO_4^{2-} \) as \( x \) and \( y \) respectively.

### Step 2: Write the Expression for Molar Conductivity
The molar conductivity (\( \Lambda_m \)) of \( H_2SO_4 \) can be expressed as the sum of the contributions from the individual ions. Since \( H_2SO_4 \) dissociates into 2 \( H^+ \) ions and 1 \( SO_4^{2-} \) ion, the molar conductivity can be written as:

\[ \Lambda_m = 2 \cdot \text{conductance of } H^+ + \text{conductance of } SO_4^{2-} \]

Substituting the values, we have:

\[ \Lambda_m = 2x + y \]

### Step 3: Calculate the Equivalent Conductivity
To find the equivalent conductivity (\( \Lambda_{eq} \)), we need to divide the molar conductivity by the number of equivalents. For \( H_2SO_4 \), which produces 2 equivalents (2 \( H^+ \)), we divide by 2:

\[ \Lambda_{eq} = \frac{\Lambda_m}{n} \]

where \( n \) is the number of equivalents. Here, \( n = 2 \):

\[ \Lambda_{eq} = \frac{2x + y}{2} \]

### Final Answer
Thus, the equivalent conductivity of \( H_2SO_4 \) is:

\[ \Lambda_{eq} = \frac{2x + y}{2} \]