The reaction of sulphur in alkaline medium is the below:
`S_(8(s))+aOH_(aq)^(-)rarrbS_((aq))^(2-)+cS_2O_(3" "aq)^(2-)+dH_2O_((l))`
The values of 'a' is `"____"`. (Integer answer)

Answer

To solve the problem, we need to balance the given chemical reaction involving sulfur in an alkaline medium. The reaction is:

\[ S_{8(s)} + aOH^{-} \rightarrow bS_{(aq)}^{2-} + cS_{2}O_{3}^{2-} + dH_{2}O_{(l)} \]

### Step-by-Step Solution:

1. **Identify Oxidation States**:
- Sulfur in \( S_8 \) has an oxidation state of 0.
- In \( S^{2-} \), the oxidation state is -2.
- In \( S_2O_3^{2-} \), we need to calculate the oxidation state of sulfur.

2. **Calculate Oxidation State of Sulfur in \( S_2O_3^{2-} \)**:
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2x + 3(-2) = -2 \]
\[ 2x - 6 = -2 \implies 2x = 4 \implies x = 2 \]
- Thus, in \( S_2O_3^{2-} \), sulfur has an oxidation state of +2.

3. **Write Half-Reactions**:
- **Reduction Half-Reaction**:
\[ S_8 + 16e^{-} \rightarrow 8S^{2-} \]
- **Oxidation Half-Reaction**:
\[ S + 2OH^{-} \rightarrow S_2O_3^{2-} + H_2O + 2e^{-} \]

4. **Balance Electrons**:
- To balance the electrons, we need to multiply the oxidation half-reaction by 8:
\[ 8S + 16OH^{-} \rightarrow 8S_2O_3^{2-} + 8H_2O + 16e^{-} \]

5. **Combine Half-Reactions**:
- Combine the reduction and oxidation half-reactions:
\[ S_8 + 16OH^{-} \rightarrow 8S^{2-} + 8S_2O_3^{2-} + 8H_2O \]

6. **Simplify the Equation**:
- Divide the entire equation by 2 to simplify:
\[ S_8 + 8OH^{-} \rightarrow 4S^{2-} + 4S_2O_3^{2-} + 4H_2O \]

7. **Identify Coefficients**:
- From the balanced equation, we can identify:
- \( a = 8 \)
- \( b = 4 \)
- \( c = 4 \)
- \( d = 4 \)

### Final Answer:
The value of \( a \) is **8**.