When `1.5mol` of `CuCI_(2).2H_(2)O` is dissolved in enough water to make `1.0L` of solution.
Given: `K_(f)CuCI^(Theta) 1.0 (K_(f)` is the formation constant of `CuCi^(o+))`
`[CiCI^(o+)]` in solution is

Options

A. `1.0 M`
B. `2.0 M`
C. `3.0 M`
D. `0.5 M`

Explanation

To solve the problem step by step, we need to find the concentration of the \( \text{CuCl}^+ \) ion in the solution after dissolving \( 1.5 \, \text{mol} \) of \( \text{CuCl}_2 \cdot 2\text{H}_2\text{O} \) in \( 1.0 \, \text{L} \) of water. The formation constant \( K_f \) for \( \text{CuCl}^+ \) is given as \( 1.0 \).

### Step 1: Determine the dissociation of \( \text{CuCl}_2 \)
When \( \text{CuCl}_2 \cdot 2\text{H}_2\text{O} \) dissolves in water, it dissociates as follows:
\[ \text{CuCl}_2 \cdot 2\text{H}_2\text{O} \rightarrow \text{Cu}^{2+} + 2\text{Cl}^- + 2\text{H}_2\text{O} \]
From \( 1.5 \, \text{mol} \) of \( \text{CuCl}_2 \), we get:
- \( 1.5 \, \text{mol} \) of \( \text{Cu}^{2+} \)
- \( 3.0 \, \text{mol} \) of \( \text{Cl}^- \)

### Step 2: Set up the equilibrium expression
The formation of \( \text{CuCl}^+ \) from \( \text{Cu}^{2+} \) and \( \text{Cl}^- \) can be represented as:
\[ \text{Cu}^{2+} + \text{Cl}^- \rightleftharpoons \text{CuCl}^+ \]
Let \( x \) be the concentration of \( \text{CuCl}^+ \) formed at equilibrium.

### Step 3: Write the equilibrium concentrations
At equilibrium:
- Concentration of \( \text{Cu}^{2+} = 1.5 - x \)
- Concentration of \( \text{Cl}^- = 3.0 - x \)
- Concentration of \( \text{CuCl}^+ = x \)

### Step 4: Write the expression for the formation constant \( K_f \)
The formation constant \( K_f \) is given by:
\[ K_f = \frac{[\text{CuCl}^+]}{[\text{Cu}^{2+}][\text{Cl}^-]} \]
Substituting the equilibrium concentrations:
\[ 1.0 = \frac{x}{(1.5 - x)(3.0 - x)} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ x = (1.5 - x)(3.0 - x) \]
Expanding the right side:
\[ x = 4.5 - 1.5x - 3.0x + x^2 \]
Rearranging gives:
\[ x^2 - 6x + 4.5 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -6, c = 4.5 \):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 4.5}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 18}}{2} = \frac{6 \pm \sqrt{18}}{2} \]
\[ x = \frac{6 \pm 3\sqrt{2}}{2} = 3 \pm \frac{3\sqrt{2}}{2} \]

### Step 7: Calculate the possible values of \( x \)
Calculating the approximate values:
1. \( x_1 = 3 + \frac{3\sqrt{2}}{2} \approx 6.12 \) (not possible since it exceeds initial concentrations)
2. \( x_2 = 3 - \frac{3\sqrt{2}}{2} \approx 0.83 \)

### Step 8: Conclusion
The concentration of \( \text{CuCl}^+ \) in the solution is approximately \( 0.83 \, \text{mol/L} \).