Question

Approximately \(0.14 \mathrm{~g}\) nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\), dissolves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Short answer

The solubility product constant, \(K_{sp}\), for nickel(II) hydroxide at 20°C is \(1.37 \times 10^{-14}\).

Step-by-step solution

Write the balanced dissolution equation

For the dissolution of nickel(II) hydroxide, the balanced equation is: \[Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)\]

Convert the solubility to molarity

We are given that 0.14 g of \(Ni(OH)_2\) dissolves per liter of water. To find the molar concentration, we will divide the mass by the molar mass of \(Ni(OH)_2\). The molar mass of \(Ni(OH)_2\) is: \(58.6934 (Ni) + 2(16.00 (O) + 1.00784 (H)) = 92.70868 \mathrm{~g/mol}\) Now, we can find the molar concentration: \[\frac{0.14 \cancel{\mathrm{g}}}{92.70868 \frac{\cancel{\mathrm{g}}}{\mathrm{mol}}} = 0.001510 \mathrm{~mol/L}\]

Determine the concentration of products

Using the stoichiometry of the balanced equation, we have: 1 mol of \(Ni(OH)_2\) dissociates into 1 mol of \(Ni^{2+}\) and 2 mol of \(OH^-\). From the molar concentration we found in Step 2, we know that 0.001510 mol of \(Ni(OH)_2\) has dissolved in 1 L of water. Therefore, we obtain the following concentrations: For \(Ni^{2+}\), \([Ni^{2+}] = 0.001510 \mathrm{~mol/L}\) For \(OH^-\), \([OH^-] = 2 \times 0.001510 = 0.003020 \mathrm{~mol/L}\)

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

Now that we have the concentrations of the products, we can use these values to calculate the \(K_{sp}\) for the dissolution of \(Ni(OH)_2\). According to the balanced equation, the \(K_{sp}\) expression is: \[K_{sp} = [Ni^{2+}][OH^-]^2\] Plugging in the concentration values from Step 3, we find: \[K_{sp} = (0.001510)(0.003020)^2 = 1.37 \times 10^{-14}\] Thus, the solubility product constant, \(K_{sp}\), for nickel(II) hydroxide at 20°C is \(1.37 \times 10^{-14}\).

Key concepts

Dissolution Equation

Understanding how solid substances dissolve in water can be tricky, but the dissolution equation simplifies this process into something much more understandable. For substances like nickel(II) hydroxide, the dissolution equation represents the chemical change from a solid to its constituent ions in solution.

For nickel(II) hydroxide, represented chemically as \(Ni(OH)_2(s)\), the dissolution equation is written as:
\[Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)\]
This tells us that for every one formula unit of \(Ni(OH)_2\) that dissolves, we will get one ion of \(Ni^{2+}\) and two hydroxide ions \(OH^-\) released into the solution. It is important to understand that this process is reversible and can reach an equilibrium state—a key concept in solubility which leads us to the solubility product constant.

Molar Concentration

Often seen as a daunting topic, molar concentration actually involves a simple calculation: the amount of substance per unit volume of solution. It's crucial for understanding the concentration of each ion in a solubility context. Here's what you need to know:

Molar concentration is typically expressed in moles per liter (mol/L) and, for the dissolution of nickel(II) hydroxide, can be found by dividing the mass of the dissolved substance by its molar mass. The calculation goes as follows:
\[\frac{mass \, of \, Ni(OH)_2}{molar \, mass \, of \, Ni(OH)_2} = \frac{0.14 \, g}{92.70868 \, g/mol} = 0.001510 \, mol/L\]
As noted in the exercise, getting a grasp of these fundamental calculations will significantly help in understanding the overall concept of solubility product constant. Remember, for soluble substances, molar concentration lets you predict how much can dissolve.

Stoichiometry

Stoichiometry plays a pivotal role in chemistry, syncing the quantitative relationships seen in chemical equations with actual measurable substances. It's all about the numbers and proportions of molecules and atoms during chemical reactions.

In context of solubility, stoichiometry suggests that for each mole of \(Ni(OH)_2\) that dissolves, we end up with a stoichiometric amount of products: one mole of \(Ni^{2+}\) and two moles of \(OH^-\). It becomes a simple mathematics game once you understand the stoichiometric coefficients from the balance dissolution equation. So, knowledge of stoichiometry is fundamental to accurately calculate the concentration of ions in solution, which ultimately allows us to find the solubility product constant.

Equilibrium Constant

In chemistry, the concept of an equilibrium constant is akin to a peace treaty between reacting substances, denoting a state where the forward and reverse reaction rates are equal. For a dissolution reaction, the equilibrium constant is particularly specific and is known as the solubility product constant (\(K_{sp}\)).

The \(K_{sp}\) is indicative of a saturated solution and is calculated by the product of the concentrations of the resulting ions, each raised to the power of their stoichiometric coefficients. For nickel(II) hydroxide, the formula looks like this:
\[K_{sp} = [Ni^{2+}][OH^-]^2\]
It's fascinating how this single constant can tell us how much of a compound will dissolve in water at a specific temperature. When you understand equilibrium constants, you can predict the extent of dissolution and even compare the solubility of different substances.