The conductivity of a 0.05 M solution of a weak monobasic acid is `10^(-3) S cm^(-1)`, If `lambda_(m)^(oo)` for weak acid `500 Scm^(2) "mol"^(-1)`, calculate Ka of weak monobasic acid :

Options

A. `8 xx 10^(-5)`
B. `4xx 10^(-6)`
C. `16 xx 10^(-7)`
D. `14 xx 10^(-8)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have:
- Conductivity (κ) of the solution = \(10^{-3} \, S \, cm^{-1}\)
- Concentration (C) of the solution = \(0.05 \, M\)
- Molar conductivity at infinite dilution (\(λ_m^{\infty}\)) = \(500 \, S \, cm^2 \, mol^{-1}\)

### Step 2: Calculate the molar conductivity at the given concentration (\(λ_{m,c}\))
The molar conductivity at the given concentration can be calculated using the formula:
\[ λ_{m,c} = \frac{1000 \cdot κ}{C} \]
Substituting the values:
\[ λ_{m,c} = \frac{1000 \cdot 10^{-3}}{0.05} \]
\[ λ_{m,c} = \frac{1}{0.05} = 20 \, S \, cm^2 \, mol^{-1} \]

### Step 3: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ α = \frac{λ_{m,c}}{λ_m^{\infty}} \]
Substituting the values:
\[ α = \frac{20}{500} \]
\[ α = 0.04 \]

### Step 4: Calculate the dissociation constant (Ka)
The dissociation constant (\(K_a\)) for the weak acid can be calculated using the formula:
\[ K_a = C \cdot α^2 \]
Substituting the values:
\[ K_a = 0.05 \cdot (0.04)^2 \]
\[ K_a = 0.05 \cdot 0.0016 \]
\[ K_a = 0.00008 = 8 \times 10^{-5} \]

### Final Answer
Thus, the dissociation constant \(K_a\) of the weak monobasic acid is \(8 \times 10^{-5}\).

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