Calculate `K_(f)` for the reaction:
`M^(3+) + SCN^(Theta) hArr MSCN^(2+)`,
The `[M^(3+)]` in the solution is `2.0 xx 10^(-3)M, [SCN^(Theta)] = 1.5 xx 10^(-3)M` and Free `[SCN^(Theta)] = 1.0 xx 10^(-5)M`.

Explanation

To calculate the formation constant \( K_f \) for the reaction:

\[ M^{3+} + SCN^{-} \rightleftharpoons MSCN^{2+} \]

we will follow these steps:

### Step 1: Write the expression for \( K_f \)

The formation constant \( K_f \) for the reaction can be expressed as:

\[ K_f = \frac{[MSCN^{2+}]}{[M^{3+}][SCN^{-}]} \]

### Step 2: Identify the concentrations

From the problem statement, we have the following concentrations:
- \([M^{3+}] = 2.0 \times 10^{-3} \, M\)
- \([SCN^{-}]_{total} = 1.5 \times 10^{-3} \, M\)
- \([SCN^{-}]_{free} = 1.0 \times 10^{-5} \, M\)

### Step 3: Calculate the concentration of \( SCN^{-} \) that forms the complex

The concentration of \( SCN^{-} \) that participates in the formation of the complex \( MSCN^{2+} \) is given by:

\[ [SCN^{-}]_{complex} = [SCN^{-}]_{total} - [SCN^{-}]_{free} \]

Substituting the values:

\[ [SCN^{-}]_{complex} = 1.5 \times 10^{-3} - 1.0 \times 10^{-5} = 1.49 \times 10^{-3} \, M \]

### Step 4: Calculate the concentration of \( M^{3+} \) remaining in solution

The concentration of \( M^{3+} \) that remains in solution after some of it has reacted to form the complex is:

\[ [M^{3+}]_{remaining} = [M^{3+}]_{initial} - [SCN^{-}]_{complex} \]

Substituting the values:

\[ [M^{3+}]_{remaining} = 2.0 \times 10^{-3} - 1.49 \times 10^{-3} = 0.51 \times 10^{-3} \, M \]

### Step 5: Substitute into the \( K_f \) expression

Now we can substitute the calculated concentrations into the \( K_f \) expression:

\[ K_f = \frac{[MSCN^{2+}]}{[M^{3+}][SCN^{-}]} \]

Since 1 mole of \( SCN^{-} \) gives 1 mole of \( MSCN^{2+} \), we have:

\[ [MSCN^{2+}] = [SCN^{-}]_{complex} = 1.49 \times 10^{-3} \, M \]

Now substituting into the \( K_f \) expression:

\[ K_f = \frac{1.49 \times 10^{-3}}{(0.51 \times 10^{-3})(1.0 \times 10^{-5})} \]

### Step 6: Calculate \( K_f \)

Calculating the above expression:

\[ K_f = \frac{1.49 \times 10^{-3}}{0.51 \times 10^{-3} \times 1.0 \times 10^{-5}} = \frac{1.49}{0.51 \times 10^{-2}} = \frac{1.49}{0.51} \times 10^{2} \]

Calculating \( \frac{1.49}{0.51} \):

\[ \frac{1.49}{0.51} \approx 2.92 \implies K_f \approx 2.92 \times 10^{5} \]

### Final Answer

Thus, the formation constant \( K_f \) for the reaction is approximately:

\[ K_f \approx 2.9 \times 10^{5} \]