Equal volumes of 0.002 M solution of sodium iodate and copper chlorate are mixed together . Will it lead to precipitation of copper iodate ? For copper iodate `K_(sp)=7.4xx10^(-4)`

Explanation

To determine whether the mixing of equal volumes of 0.002 M sodium iodate and copper chlorate will lead to the precipitation of copper iodate, we can follow these steps:

### Step 1: Write the dissociation equations
1. Sodium iodate (NaIO3) dissociates into sodium ions (Na⁺) and iodate ions (IO3⁻):
\[ \text{NaIO}_3 \rightarrow \text{Na}^+ + \text{IO}_3^- \]

2. Copper chlorate (Cu(ClO3)2) dissociates into copper ions (Cu²⁺) and chlorate ions (ClO3⁻):
\[ \text{Cu(ClO}_3)_2 \rightarrow \text{Cu}^{2+} + 2\text{ClO}_3^- \]

### Step 2: Calculate the concentrations after mixing
Since equal volumes of both solutions are mixed, the total volume doubles. Therefore, the concentration of each ion after mixing will be halved.

- Initial concentration of NaIO3 = 0.002 M
- After mixing:
\[ \text{Concentration of IO}_3^- = \frac{0.002}{2} = 0.001 \, \text{M} \]

- Initial concentration of Cu(ClO3)2 = 0.002 M
- After mixing:
\[ \text{Concentration of Cu}^{2+} = \frac{0.002}{2} = 0.001 \, \text{M} \]

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) expression for copper iodate (Cu(IO3)2) is given by:
\[ K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^-]^2 \]

### Step 4: Calculate the ionic product (IP)
Substituting the concentrations into the Ksp expression:
\[ IP = [\text{Cu}^{2+}][\text{IO}_3^-]^2 = (0.001)(0.001)^2 = 0.001 \times 0.000001 = 1 \times 10^{-9} \]

### Step 5: Compare IP with Ksp
Given:
\[ K_{sp} = 7.4 \times 10^{-4} \]
Now we compare:
\[ IP = 1 \times 10^{-9} \quad \text{and} \quad K_{sp} = 7.4 \times 10^{-4} \]

Since \( IP < K_{sp} \), there will be no precipitation.

### Conclusion
No precipitation of copper iodate will occur when equal volumes of 0.002 M sodium iodate and copper chlorate are mixed together.

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