If the concentration of` Y ^(-)` ion in the reaction
`XY_(3(s)) hArr X_(aq)^(+3)+3Y^(-)(aq)` is
decreased by `(1)/(2)` times, then equilibrium `conc^(n)` of
`X^(+3)` will increase by :-

Options

A. 2 times
B. 4 times
C. 8 times
D. 16 times

Explanation

To solve the problem, we need to analyze the equilibrium reaction and how the change in concentration of the `Y^(-)` ion affects the concentration of `X^(+3)` ion.

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction**:
The given reaction is:
\[ XY_3(s) \rightleftharpoons X^{+3}(aq) + 3Y^{-}(aq) \]

2. **Write the Expression for Equilibrium Constant (Kc)**:
The equilibrium constant expression for this reaction is given by:
\[ K_c = \frac{[X^{+3}]}{[Y^{-}]^3} \]
Here, `[X^{+3}]` is the concentration of `X^(+3)` ions and `[Y^{-}]` is the concentration of `Y^(-)` ions.

3. **Initial Concentration of Y^-**:
Let the initial concentration of `Y^-` be `[Y^{-}] = Y_0`.

4. **Decrease Concentration of Y^-**:
According to the problem, the concentration of `Y^-` is decreased by half:
\[ [Y^{-}] = \frac{1}{2} Y_0 \]

5. **Substituting into the Kc Expression**:
Now we substitute the new concentration of `Y^-` into the Kc expression:
\[ K_c = \frac{[X^{+3}]}{(\frac{1}{2} Y_0)^3} \]
Simplifying this gives:
\[ K_c = \frac{[X^{+3}]}{\frac{1}{8} Y_0^3} = 8 \cdot \frac{[X^{+3}]}{Y_0^3} \]

6. **Setting the Two Kc Expressions Equal**:
Let’s denote the initial concentration of `X^(+3)` as `[X^{+3}] = X_0`. Therefore, we have:
\[ K_c = \frac{X_0}{Y_0^3} \]
Setting the two expressions for `K_c` equal gives:
\[ \frac{X_0}{Y_0^3} = 8 \cdot \frac{[X^{+3}]}{Y_0^3} \]

7. **Solving for New Concentration of X^(+3)**:
Canceling out `Y_0^3` from both sides:
\[ X_0 = 8 \cdot [X^{+3}] \]
Therefore, the new concentration of `X^(+3)` can be expressed as:
\[ [X^{+3}] = \frac{X_0}{8} \]

8. **Conclusion**:
The equilibrium concentration of `X^(+3)` will increase by a factor of 8 when the concentration of `Y^(-)` is decreased by half.

### Final Answer:
The equilibrium concentration of `X^(+3)` will increase by a factor of **8**.