For a weak electrolyte `alpha_(1) and alpha_(2)` are in ratio of `1:2`, for a given concentration `k_(a_(1))=2xx10^(-4)`. What will be value of `k_(a_(2))`?

Options

A. `8xx10^(-4)`
B. `2xx10^(-4)`
C. `4xx10^(-4)`
D. `1xx10^(-4)`

Answer

To solve the problem, we need to find the value of \( k_{a_2} \) given the following information:

1. The ratio of the dissociation constants \( \alpha_1 \) and \( \alpha_2 \) is \( 1:2 \).
2. The value of \( k_{a_1} \) is \( 2 \times 10^{-4} \).

### Step-by-step Solution:

**Step 1: Write down the given ratios and values.**
- We have \( \frac{\alpha_1}{\alpha_2} = \frac{1}{2} \).
- We know \( k_{a_1} = 2 \times 10^{-4} \).

**Step 2: Use the relationship between dissociation constants.**
- The relationship between the dissociation constants and the degree of dissociation is given by:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{k_{a_1}}{k_{a_2}}} \]

**Step 3: Substitute the known values into the equation.**
- Plugging in the values we have:
\[ \frac{1}{2} = \sqrt{\frac{2 \times 10^{-4}}{k_{a_2}}} \]

**Step 4: Square both sides to eliminate the square root.**
- Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{2 \times 10^{-4}}{k_{a_2}} \]
\[ \frac{1}{4} = \frac{2 \times 10^{-4}}{k_{a_2}} \]

**Step 5: Rearrange the equation to solve for \( k_{a_2} \).**
- Rearranging gives:
\[ k_{a_2} = 2 \times 10^{-4} \times 4 \]

**Step 6: Calculate \( k_{a_2} \).**
- Performing the multiplication:
\[ k_{a_2} = 8 \times 10^{-4} \]

### Final Answer:
The value of \( k_{a_2} \) is \( 8 \times 10^{-4} \).

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