In an aqueous solution of volume `500 ml` when the reaction `2Ag^(+)(aq)+Cu(s) hArr Cu^(2+)(aq)+2Ag(s)` reached equilibrium, the `[Cu^(2+)]` was `'a'M`. If `500 ml` water is further added, at the equilibrium `[Cu^(2+)]` will be `:`

Options

A. `'a'//2M`
B. `'a'M`
C. between `'a'` and `'a//2'M`
D. less than `'a//2'`

Explanation

To solve the problem, we need to determine the new concentration of \([Cu^{2+}]\) after adding 500 ml of water to the original 500 ml solution, where the concentration of \([Cu^{2+}]\) was initially \(a\) M.

### Step-by-Step Solution:

1. **Initial Concentration**: The initial concentration of \([Cu^{2+}]\) is given as \(a\) M in a volume of 500 ml.

2. **Calculate Moles of \(Cu^{2+}\)**:
- The number of moles of \(Cu^{2+}\) can be calculated using the formula:
\[ \text{Moles of } Cu^{2+} = \text{Concentration} \times \text{Volume} = a \times 0.5 \, \text{L} = 0.5a \, \text{moles} \]

3. **New Volume After Adding Water**:
- When we add 500 ml of water, the total volume becomes:
\[ \text{Total Volume} = 500 \, \text{ml} + 500 \, \text{ml} = 1000 \, \text{ml} = 1 \, \text{L} \]

4. **Calculate New Concentration**:
- The concentration of \(Cu^{2+}\) after the volume change can be calculated using the formula:
\[ \text{New Concentration} = \frac{\text{Moles of } Cu^{2+}}{\text{Total Volume}} = \frac{0.5a}{1 \, \text{L}} = 0.5a \, \text{M} \]

5. **Final Answer**:
- Therefore, after adding 500 ml of water, the new concentration of \([Cu^{2+}]\) will be:
\[ \frac{a}{2} \, \text{M} \]

### Summary:
The new concentration of \([Cu^{2+}]\) after adding 500 ml of water will be \(\frac{a}{2}\) M.