At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`. The ionic conductances of `H^(+),C_(2)H_(5)O^(-)` at his temperature is 300 and 100 `mhocm^(2)" equivalent"^(-1)` respectively. Then the negative logarithm of ionic product of alcohol will be 18.

Explanation

To solve the problem step by step, we will follow the logical flow of the information provided:

### Step 1: Understand the Given Data
- Specific conductivity of ethanol (K) = \(4 \times 10^{-10} \, \text{mho cm}^{-1}\)
- Ionic conductance of \(H^+\) = 300 mho cm² equivalent⁻¹
- Ionic conductance of \(C_2H_5O^-\) = 100 mho cm² equivalent⁻¹

### Step 2: Calculate the Equivalent Conductance at Infinite Dilution
The equivalent conductance at infinite dilution (\(\Lambda\)) is the sum of the ionic conductances of the ions present in the solution:
\[ \Lambda = \Lambda_{H^+} + \Lambda_{C_2H_5O^-} \]
Substituting the given values:
\[ \Lambda = 300 + 100 = 400 \, \text{cm}^2 \, \text{equivalent}^{-1} \]

### Step 3: Relate Specific Conductivity to Concentration
The relationship between specific conductivity (K), equivalent conductance (\(\Lambda\)), and concentration (C) is given by the formula:
\[ K = \Lambda \times C \]
We can rearrange this to find the concentration:
\[ C = \frac{K}{\Lambda} \]

### Step 4: Substitute the Values to Find Concentration
Substituting the values of K and \(\Lambda\):
\[ C = \frac{4 \times 10^{-10} \, \text{mho cm}^{-1}}{400 \, \text{cm}^2 \, \text{equivalent}^{-1}} \]
Calculating this gives:
\[ C = \frac{4 \times 10^{-10}}{400} = 1 \times 10^{-12} \, \text{equivalent cm}^{-3} \]
Since we want the concentration in molarity (M), we convert equivalent cm³ to liters:
\[ C = 1 \times 10^{-12} \, \text{equivalent cm}^{-3} = 1 \times 10^{-9} \, \text{M} \]

### Step 5: Calculate the Ionic Product of Alcohol
The ionic product (\(K_{alcohol}\)) can be calculated using the concentration:
\[ K_{alcohol} = [H^+][C_2H_5O^-] = C^2 = (1 \times 10^{-9})^2 = 1 \times 10^{-18} \]

### Step 6: Calculate the Negative Logarithm of the Ionic Product
To find the negative logarithm of the ionic product:
\[ pK_{alcohol} = -\log(K_{alcohol}) = -\log(1 \times 10^{-18}) = 18 \]

### Final Answer
The negative logarithm of the ionic product of alcohol is 18.

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