Question

Zinc carbonate, \(\mathrm{ZnCO}_{3}(\mathrm{s}),\) dissolves in water to give a solution that is \(1.7 \times 10^{-5} \mathrm{M}\) at \(22^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{ZnCO}_{3}(s)\) at this temperature..

Short answer

The solubility product constant, \(K_{\mathrm{sp}}\), for zinc carbonate, \(\mathrm{ZnCO}_{3}(\mathrm{s})\), at \(22^{\circ} \mathrm{C}\) is approximately \(2.89 \times 10^{-10}\).

Step-by-step solution

Write the balanced dissolution equation for \(\mathrm{ZnCO}_{3}(\mathrm{s})\)

Write the balanced chemical equation for the dissolution of zinc carbonate, \(\mathrm{ZnCO}_{3}(\mathrm{s})\), in water: \[\mathrm{ZnCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Zn}^{2+} (\mathrm{aq}) + \mathrm{CO}_{3}^{2-} (\mathrm{aq})\]

Use the given molar concentration to determine ion concentrations

We are given the molar concentration of the resulting solution to be \(1.7 \times 10^{-5} \mathrm{M}\). Since the balanced equation shows a 1:1 ratio of \(\mathrm{Zn}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\) ions, we can conclude that the concentrations of both ions are equal and equal to \(1.7 \times 10^{-5} \mathrm{M}\): \[[\mathrm{Zn}^{2+}] = [\mathrm{CO}_{3}^{2-}] = 1.7 \times 10^{-5} \mathrm{M}\]

Calculate the solubility product constant, \(K_{\mathrm{sp}}\)

Using the balanced dissolution equation, we can write the expression for the solubility product constant, \(K_{\mathrm{sp}}\): \[K_{\mathrm{sp}} = [\mathrm{Zn}^{2+}][\mathrm{CO}_{3}^{2-}]\] Now, we can plug in the concentrations we found in Step 2: \[K_{\mathrm{sp}} = (1.7 \times 10^{-5})(1.7 \times 10^{-5})\]

Compute the value of \(K_{\mathrm{sp}}\)

Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{ZnCO}_{3}(\mathrm{s})\): \[K_{\mathrm{sp}} = (1.7 \times 10^{-5})^2 = 2.89 \times 10^{-10}\] So, the solubility product constant for zinc carbonate at \(22^{\circ} \mathrm{C}\) is approximately \(2.89 \times 10^{-10}\).

Key concepts

Zinc Carbonate Dissolution

When zinc carbonate, written as \( \mathrm{ZnCO}_{3}(\mathrm{s}) \), dissolves in water, it breaks down into ions. This is the process of dissolution. In simple terms, the solid zinc carbonate separates into its ionic components when it comes into contact with water. Zinc carbonate dissociates according to this balanced chemical equation:\[ \mathrm{ZnCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Zn}^{2+} (\mathrm{aq}) + \mathrm{CO}_{3}^{2-} (\mathrm{aq}) \]- \( \mathrm{Zn}^{2+} \) is a zinc ion.- \( \mathrm{CO}_{3}^{2-} \) is a carbonate ion.The arrow in the equation represents a reversible process, which means zinc carbonate can break down into ions and also recombine back into the solid form under certain conditions. This step forms the basis for understanding how substances dissolve in a solvent and helps visualize the process occurring in solution.

Chemical Equilibrium

When we discuss dissolving zinc carbonate, we enter the realm of chemical equilibrium. At this stage, the rate at which zinc carbonate dissolves is equal to the rate at which the ions recombine to form the solid. Chemical equilibrium in a solution is like a seesaw perfectly balanced, where reactions in both directions occur at the same rate. - At equilibrium, there are no net changes in the concentration of the dissolved ions or the undissolved solid. - This equilibrium matters because it helps us predict how much of a compound will dissolve in the water under specific conditions. It's important to understand that even when at equilibrium, zinc carbonate does not dissolve completely. Only a tiny amount becomes ions, leading to the existence of a dynamic balance in the mixture.

Molar Concentration

The molar concentration tells us how much of a substance is present in a specific volume of solution. In this exercise, the important concentration is of the dissolved zinc carbonate.Given as \( 1.7 \times 10^{-5} \mathrm{M} \), it tells us the molarity, which is the number of moles of substance per liter of solution. - Since \( \mathrm{Zn}^{2+} \) and \( \mathrm{CO}_{3}^{2-} \) ions form a 1:1 ratio, their concentrations in the solution are equal, each being \( 1.7 \times 10^{-5} \mathrm{M} \). Molar concentration is very useful because it allows us to understand the extent of dissolution and is a critical part of calculating the solubility product constant.

Ion Concentration Calculation

To calculate ion concentrations in a solution, look at the balanced chemical equation and the molar concentration given. The equation for zinc carbonate dissociation provides a straightforward guide:\[ \mathrm{ZnCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Zn}^{2+} (\mathrm{aq}) + \mathrm{CO}_{3}^{2-} (\mathrm{aq}) \]From this, we understand that for every formula unit of zinc carbonate that dissolves: - One zinc ion \( \mathrm{Zn}^{2+} \) is formed - One carbonate ion \( \mathrm{CO}_{3}^{2-} \)Both ions will have the same concentration when solution equilibrium is reached. Here, the concentration is \( 1.7 \times 10^{-5} \mathrm{M} \). Multiply these concentrations to find the solubility product constant (\( K_{\mathrm{sp}} \)): \[ K_{\mathrm{sp}} = [\mathrm{Zn}^{2+}] \times [\mathrm{CO}_{3}^{2-}] = (1.7 \times 10^{-5})^2 = 2.89 \times 10^{-10} \]This value helps predict the solubility behavior of zinc carbonate under these conditions.