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Equal volumes of 0.002 M solution of sod...

Equal volumes of 0.002 M solution of sodium iodate and cupric chlorate are mixed togather. Will it lead to precipitation of copper iodate?
`("for cupric iodate" K =7.4xx10^(-8))`.

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To determine whether precipitation of copper iodate will occur when equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed, we can follow these steps: ### Step 1: Understand the Reaction When sodium iodate (NaIO3) is mixed with cupric chlorate (Cu(ClO3)2), the reaction can be represented as: \[ \text{Cu}^{2+} + 2 \text{IO}_3^{-} \rightarrow \text{Cu(IO}_3\text{)}_2 \] We need to check if copper iodate (Cu(IO3)2) will precipitate out of the solution. ### Step 2: Calculate Initial Concentrations Given that both solutions are 0.002 M and equal volumes are mixed, the final volume after mixing will be double the original volume (2V). The concentrations of the ions after mixing will be halved: - Concentration of \(\text{Cu}^{2+}\) = \( \frac{0.002 \, \text{M}}{2} = 0.001 \, \text{M} \) - Concentration of \(\text{IO}_3^{-}\) = \( \frac{0.002 \, \text{M}}{2} = 0.001 \, \text{M} \) ### Step 3: Write the Expression for Ksp The solubility product constant (Ksp) for copper iodate is given as \( K_{sp} = 7.4 \times 10^{-8} \). The expression for Ksp for the precipitation of copper iodate can be written as: \[ K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^{-}]^2 \] ### Step 4: Calculate the Ionic Product (Q) Now, we can calculate the ionic product (Q) using the concentrations obtained: \[ Q = [\text{Cu}^{2+}][\text{IO}_3^{-}]^2 = (0.001)(0.001)^2 = 0.001 \times 0.000001 = 1 \times 10^{-9} \] ### Step 5: Compare Q with Ksp Now we compare the ionic product (Q) with the solubility product (Ksp): - \( Q = 1 \times 10^{-9} \) - \( K_{sp} = 7.4 \times 10^{-8} \) Since \( Q < K_{sp} \), this indicates that the solution is unsaturated with respect to copper iodate. ### Conclusion Since the ionic product (Q) is less than the solubility product (Ksp), no precipitation of copper iodate will occur when equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed. ---

To determine whether precipitation of copper iodate will occur when equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed, we can follow these steps: ### Step 1: Understand the Reaction When sodium iodate (NaIO3) is mixed with cupric chlorate (Cu(ClO3)2), the reaction can be represented as: \[ \text{Cu}^{2+} + 2 \text{IO}_3^{-} \rightarrow \text{Cu(IO}_3\text{)}_2 \] We need to check if copper iodate (Cu(IO3)2) will precipitate out of the solution. ### Step 2: Calculate Initial Concentrations ...
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