Question

Given \(K_{s p}\) and the equilibrium concentration of one ion, calculate the equilibrium concentration of the other ion. (a) cadmium(II) hydroxide: \(K_{\text {sp }}=2.5 \times 10^{-14} ;\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M\) (b) copper(II) arsenate \(\left(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right): K_{\mathrm{sp}}=7.6 \times 10^{-36} ;\left[\mathrm{AsO}_{4}^{3-}\right]=\) \(2.4 \times 10^{-4} M\) (c) zinc oxalate: \(K_{s p}=2.7 \times 10^{-8} ;\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.8 \times 10^{-3} M\)

Short answer

Question: Calculate the equilibrium concentration of ions for (a) hydroxide ions when the equilibrium concentration of cadmium(II) ions is \(1.5 \times 10^{-6} M\) and the \(K_{sp}\) for cadmium(II) hydroxide is \(2.5 \times 10^{-14}\), (b) copper(II) ions when the equilibrium concentration of arsenate ions is \(2.4 \times 10^{-4} M\) and the \(K_{sp}\) for copper(II) arsenate is \(7.6 \times 10^{-36}\), and (c) zinc(II) ions when the equilibrium concentration of oxalate ions is \(8.8 \times 10^{-3} M\) and the \(K_{sp}\) for zinc oxalate is \(2.7 \times 10^{-8}\). Answer: (a) \([\mathrm{OH^-}] = 4.09 \times 10^{-5} M\), (b) \([\mathrm{Cu^{2+}}] = 1.1 \times 10^{-9} M\), (c) \([\mathrm{Zn^{2+}}] = 3.07\times10^{-6} M\)

Step-by-step solution

Write the dissociation reaction

First, write the balanced chemical equation for the dissociation of cadmium(II) hydroxide in water:$$\mathrm{Cd(OH)_{2}(s)} \leftrightharpoons \mathrm{Cd^{2+}(aq)} + 2\mathrm{OH^{-}(aq)}$$

Write the expression for Ksp

Next, write the expression for the solubility product constant, \(K_{sp}\):$$K_{\text {sp }}=\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{OH^{-}\right]^{2}$$

Substitute the given values and solve

Now we can substitute the given values for \(K_{sp}\) and \([\mathrm{Cd}^{2+}]\), and solve for \([\mathrm{OH^-}]\): $$K_{\text {sp }}=2.5 \times 10^{-14}$$ $$\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M$$ $$2.5 \times 10^{-14} = (1.5 \times 10^{-6})\left[\mathrm{OH^-}\right]^2$$ Divide both sides by \((1.5 \times 10^{-6})\): $$\left[\mathrm{OH^-}\right]^2=1.67 \times 10^{-8}$$ Now, we can take the square root of both sides: $$[\mathrm{OH^-}] = \sqrt{1.67 \times 10^{-8}} = 4.09 \times 10^{-5} M$$ For copper(II) arsenate (Cu3(AsO4)2):

Write the dissociation reaction

First, write the balanced chemical equation for the dissociation of copper(II) arsenate in water:$$\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}(s) \leftrightharpoons 3\mathrm{Cu^{2+}(aq)} + 2\mathrm{AsO}_{4}^{3-}(aq)$$

Write the expression for Ksp

Next, write the expression for the solubility product constant, \(K_{sp}\):$$K_{\text {sp }}=\left[\mathrm{Cu^{2+}}\right]^{3}\left[\mathrm{AsO_{4}^{3-}}\right]^{2}$$

Substitute the given values and solve

Now we can substitute the given values for \(K_{sp}\) and \([\mathrm{AsO_{4}^{3-}}]\), and solve for \([\mathrm{Cu^{2+}}]\): $$K_{\text {sp }}=7.6 \times 10^{-36}$$ $$[\mathrm{AsO_{4}^{3-}}]=2.4 \times 10^{-4} M$$ $$7.6 \times 10^{-36} =\left[\mathrm{Cu^{2+}}\right]^{3}(2.4 \times 10^{-4})^2$$ Divide both sides by \({(2.4 \times 10^{-4})^2}\): $$\left[\mathrm{Cu^{2+}}\right]^3 = 1.32 \times 10^{-27}$$ Now, we can take the cube root of both sides: $$[\mathrm{Cu^{2+}}] = \sqrt[3]{1.32 \times 10^{-27}} = 1.1 \times 10^{-9} M$$ For zinc oxalate (ZnC2O4):

Write the dissociation reaction

First, write the balanced chemical equation for the dissociation of zinc oxalate in water:$$\mathrm{ZnC}_{2}\mathrm{O}_{4}(s) \leftrightharpoons \mathrm{Zn^{2+}(aq)} + \mathrm{C}_{2}\mathrm{O}_{4}{ }^{2-}(aq)$$

Write the expression for Ksp

Next, write the expression for the solubility product constant, \(K_{sp}\):$$K_{\text {sp }}=\left[\mathrm{Zn^{2+}}\right]\left[\mathrm{C}_{2}\mathrm{O}_{4}{ }^{2-}\right]$$

Substitute the given values and solve

Now we can substitute the given values for \(K_{sp}\) and \([\mathrm{C}_{2}\mathrm{O}_{4}{ }^{2-}]\), and solve for \([\mathrm{Zn^{2+}}]\): $$K_{\text {sp }}=2.7 \times 10^{-8}$$ $$[\mathrm{C}_{2}\mathrm{O}_{4}{ }^{2-}]=8.8 \times 10^{-3} M$$ $$2.7 \times 10^{-8}=\left[\mathrm{Zn^{2+}}\right](8.8 \times 10^{-3})$$ Divide both sides by \(8.8 \times 10^{-3}\): $$[\mathrm{Zn^{2+}}] = 3.07\times10^{-6} M$$ In conclusion, the equilibrium ion concentrations are as follows: (a) \(\left[\mathrm{OH^{-}}\right]=4.09 \times 10^{-5} M\) (b) \(\left[\mathrm{Cu^{2+}}\right]=1.1 \times 10^{-9} M\) (c) \(\left[\mathrm{Zn^{2+}}\right]=3.07\times10^{-6} M\)

Key concepts

Equilibrium Concentration

When a chemical reaction reaches equilibrium, the concentration of reactants and products remains constant. This is because the rate at which the reactants react to form the products is equal to the rate at which the products revert to the reactants.
In the context of a solubility product problem, like our exercise, equilibrium concentration refers to the stable concentration of ions in a saturated solution of a slightly soluble salt.
For instance, after dissociation, the ions remain in balance, resulting in no net change in their concentration.

• Equilibrium concentration ensures that the dissolving and precipitating rates are equal.
• It is crucial for calculating solubility product constant (Ksp) values.

Understanding this helps predict how much of a solute can dissolve in a solution before reaching saturation.

Dissociation Reaction

The dissociation reaction is a vital step in understanding how compounds dissolve in water to form ions. In our scenario, compounds like cadmium(II) hydroxide, copper(II) arsenate, and zinc oxalate dissociate in water, splitting into their respective ions.
For example, cadmium(II) hydroxide dissociates as follows:
\[\mathrm{Cd(OH)_2(s)} \leftrightharpoons \mathrm{Cd^{2+}(aq)} + 2\mathrm{OH^{-}(aq)}\]This reaction increases the concentration of ions in the solution. It is an example of a reversible reaction, meaning it can go both ways - ions coming together or breaking apart.
The dissociation reaction is key to finding the equilibrium in the solution and calculating the solubility product constant.

Chemical Equation

A chemical equation represents the process by which reactants transform into products. It is a concise way of showing what happens in a chemical reaction, indicating the substances involved and their proportions.
For the solubility product constant (Ksp) calculations, it's crucial to write a balanced chemical equation. This helps in figuring out the exact stoichiometric relationship between different ions.

• A balanced equation is necessary for accurate Ksp expression.
• It provides the ratio of ions produced when a compound dissolves.

The balanced chemical equations in our exercise guide the calculation of equilibrium concentrations of ions, ensuring the steps are precise and reliable.

Ion Concentration

Ion concentration refers to the amount of a specific ion present in a solution. It is measured in molarity (M), which represents moles per liter.
In solubility equilibria, knowing the ion concentration is critical to solving for the solubility product constant (Ksp). For example, if given the concentration of one ion, we can calculate the other's equilibrium concentration using the Ksp equation.

• Higher ion concentration will lead to a saturated solution, potentially resulting in precipitation.
• It is calculated by rearranging the Ksp expression, solving for the unknown ion concentration.

Understanding how ion concentrations work helps predict the extent of solubility and the conditions under which a solution becomes saturated.