Given the equivalent conductance of sodium butyrate sodium chloride and hydrogen chloride as 83,127 and 426 mho `cm^(2)` at `25^(@)C` respectively the equivalent conductance of butyric acid at infinite dilution.

Explanation

To find the equivalent conductance of butyric acid at infinite dilution, we can use the given equivalent conductances of sodium butyrate, sodium chloride, and hydrogen chloride. Here's a step-by-step solution:

### Step 1: Write the equations for the given compounds
1. **Sodium Butyrate (C4H7COONa)**:
\[ \Lambda^0_{\text{C4H7COONa}} = \Lambda^0_{\text{C4H7COO}^-} + \Lambda^0_{\text{Na}^+} \]
Given: \(\Lambda^0_{\text{C4H7COONa}} = 83 \, \text{mho cm}^2\)

2. **Hydrochloric Acid (HCl)**:
\[ \Lambda^0_{\text{HCl}} = \Lambda^0_{\text{H}^+} + \Lambda^0_{\text{Cl}^-} \]
Given: \(\Lambda^0_{\text{HCl}} = 426 \, \text{mho cm}^2\)

3. **Sodium Chloride (NaCl)**:
\[ \Lambda^0_{\text{NaCl}} = \Lambda^0_{\text{Na}^+} + \Lambda^0_{\text{Cl}^-} \]
Given: \(\Lambda^0_{\text{NaCl}} = 127 \, \text{mho cm}^2\)

### Step 2: Set up the equations
From the above equations, we can express the equivalent conductance of butyric acid (C4H7COOH) as:
\[ \Lambda^0_{\text{C4H7COOH}} = \Lambda^0_{\text{C4H7COO}^-} + \Lambda^0_{\text{H}^+} \]

### Step 3: Eliminate the common ions
To find \(\Lambda^0_{\text{C4H7COO}^-} + \Lambda^0_{\text{H}^+}\), we can use the following relationship:
\[ \Lambda^0_{\text{C4H7COO}^-} + \Lambda^0_{\text{H}^+} = \Lambda^0_{\text{C4H7COONa}} + \Lambda^0_{\text{HCl}} - \Lambda^0_{\text{NaCl}} \]

### Step 4: Substitute the known values
Now, substituting the known values into the equation:
\[ \Lambda^0_{\text{C4H7COOH}} = 83 + 426 - 127 \]

### Step 5: Calculate the equivalent conductance
Calculating the above expression:
\[ \Lambda^0_{\text{C4H7COOH}} = 83 + 426 - 127 = 382 \, \text{mho cm}^2 \]

### Final Answer
The equivalent conductance of butyric acid at infinite dilution is:
\[ \Lambda^0_{\text{C4H7COOH}} = 382 \, \text{mho cm}^2 \]

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