Problem 27 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Calculate the maximum quantity (in mol) of \(\mathrm{KIO}_{3}\) that can be added to \(250 \mathrm{cm}^{3}\) of a solution containing \(1.00 \times 10^{-3} \mathrm{mol} \mathrm{dm}^{-3}\) of \(\mathrm{Cu}^{2}\) ' (aq) without precipitating \(\mathrm{Cu}(\mathrm{IO} 3)_{2}(\mathrm{s}) . K_{\mathrm{sp}}=1.4 \times 10^{-7}\) for \(\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s})\).

Step-by-Step Solution

Answer

The maximum concentration of \(\text{IO}_3^-\) is approximately \(1.183 \times 10^{-2} \text{ mol dm}^{-3}\). In 250 cm³, this is approximately \(2.96 \times 10^{-3} \text{ mol}\).

Step 1: Calculate the Solubility of Copper Iodate

For the compound \( ext{Cu(IO}_3)_2\), the solubility product is defined as \(K_{sp} = [ ext{Cu}^{2+}][ ext{IO}_3^-]^2\). We are given \(K_{sp} = 1.4 \times 10^{-7}\) and the concentration of \(\text{Cu}^{2+}\) ions as \(1.00 \times 10^{-3} \text{ mol dm}^{-3}\). Substitute these values into the \(K_{sp}\) expression to find the concentration of \([ ext{IO}_3^-]\): \[ 1.4 \times 10^{-7} = (1.00 \times 10^{-3})([ ext{IO}_3^-])^2 \] Solving for \([ ext{IO}_3^-]\) yields:\[ [ ext{IO}_3^-]^2 = \frac{1.4 \times 10^{-7}}{1.00 \times 10^{-3}} = 1.4 \times 10^{-4} \] \[ [ ext{IO}_3^-] = \sqrt{1.4 \times 10^{-4}} \approx 1.183 \times 10^{-2} \text{ mol dm}^{-3}. \]

Key Concepts

Copper Iodate

Copper iodate is a chemical compound with the formula \( \text{Cu(IO}_3)_2 \). It consists of copper ions \( \text{Cu}^{2+} \) and iodate ions \( \text{IO}_3^- \). This compound is interesting in the study of solubility because it is slightly soluble in water, making it a good candidate for exploring solubility products and chemical equilibrium.

When copper iodate is added to a solution, it typically dissociates into its ions until a balance or equilibrium is established. This is because, according to the principle of chemical equilibrium, the rate of the forward reaction (dissolving) equals the rate of the reverse reaction (precipitation). This balance is defined by its solubility product \( K_{sp} \).

Understanding copper iodate in the context of solubility helps us comprehend how compounds dissolve in solutions and how to manipulate this process, facilitating numerous applications in chemistry, including the prediction of precipitate formation in reactions.

Solubility Calculations

Solubility calculations allow us to determine how much of a compound can dissolve in a solution before full saturation and precipitation occur. This was the core process in calculating the maximum amount of potassium iodate that could be added to the solution without causing copper iodate to precipitate.

The equation used for these calculations is derived from the expression for its solubility product \( K_{sp} \). For copper iodate, this expression is \( K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^-]^2 \). By substituting known values into this formula, we can solve for the concentration of the iodate ion \( [\text{IO}_3^-] \).

• First, rearrange the \( K_{sp} \) formula to solve for \( [\text{IO}_3^-]^2 \).
• Insert the given concentration of \( \text{Cu}^{2+} \) into the equation.
• Calculate \( [\text{IO}_3^-] \) by taking the square root of the result.

This result tells us the concentration at which the solubility equilibrium is maintained and any more addition would lead to precipitation.

Chemical Equilibrium

Chemical equilibrium is a concept where the forward and reverse reactions occur at the same rate, leading to no net change in concentration of reactants and products over time. In the context of solubility and precipitation, chemical equilibrium describes the balance between dissolved ions and any solid precipitate in a saturated solution.

For copper iodate in solution, achieving chemical equilibrium means that the rate at which copper iodate dissolves is equal to the rate at which it precipitates. At this point of equilibrium, the product of the ion concentrations, raised to the power of their stoichiometric coefficients, equals the \( K_{sp} \) value.

• When considering a dynamic system like a solution, equilibrium is crucial for predicting how much of a substance can dissolve without causing excess solid to form.
• This balance ensures that no additional solid will form as long as the conditions remain unchanged, abiding by the \( K_{sp} \) rule.
• Understanding equilibrium helps in anticipating the outcomes of chemical reactions, especially when dealing with reactions that involve the formation of sparingly soluble salts.

By manipulating concentration and understanding \( K_{sp} \), one can effectively control and predict the solubility behavior of compounds within an aqueous solution.