91Ó°ÊÓ

Molarity is a measure of the concentration of a solute in a solution. It is usually expressed in moles per liter (M or mol/L). To calculate molarity, you need to know two things: the number of moles of the solute and the volume of the solution in liters.

For instance, in this problem, we have 0.1 M \(\text{NiCl}_2\) and \(250 \text{ mL}\) of solution, which translates to 0.025 moles of \(\text{NiCl}_2\). This is achieved by multiplying the molarity by the volume in liters, in this case \(0.1 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ moles}\).

Addition of \(\text{NaOH}\) changes the scenario, as it starts reacting with \(\text{NiCl}_2\). This changes the concentrations of the solution components, impacting the calculation of pH.

Key points about molarity:

• Always convert volume to liters for molarity calculations.
• Remember to account for changes in concentration when reactions occur, as seen when \(\text{Ni}^{2+}\) reacts with \(\text{OH}^-\).

Equilibrium constant (\(K_{sp}\)) is a critical concept that applies to the solubility of sparingly soluble compounds. It provides a quantitative measure of the solubility product of an ionic compound.

In this exercise, we focus on \(\text{Ni(OH)}_2\) which has a \(K_{sp}\) of \(1.6 \times 10^{-14}\). This value helps us understand how many \(\text{Ni}^{2+}\) and \(\text{OH}^-\) ions are in equilibrium in solution when dissolution and precipitation processes are happening simultaneously.

To calculate the \([\text{OH}^-]\) concentration using \(K_{sp}\), use the equation \([\text{Ni}^{2+}][\text{OH}^-]^2 = 1.6 \times 10^{-14}\). Since all \(\text{OH}^-\) typically goes to precipitate reaction, additional calculations or assumptions may be necessary.

Understanding \(K_{sp}\) is crucial when considering how precipitation affects \(pH\) and solubility.

Key points:

• \(K_{sp}\) provides a limit to how much solute can dissolve.
• In precipitating reactions, it shows the maximum concentration of ions possible before solid forms again.

In a chemical reaction, the limiting reactant is the substance that is completely consumed first and therefore limits the amount of product that can be formed.

When \(\text{NaOH}\) and \(\text{NiCl}_2\) react in this exercise, the limiting reactant can be determined by comparing the stoichiometric requirements. \(\text{Ni}^{2+}\) from \(\text{NiCl}_2\) needs two hydroxide ions (\(\text{OH}^-\)) for each ion to fully react and form \(\text{Ni(OH)}_2\) precipitate.

Here, \(0.025\) moles of \(\text{Ni}^{2+}\) require \(0.05\) moles of \(\text{OH}^-\), but there are only \(0.04375\) moles of \(\text{OH}^-\) present, making \(\text{Ni}^{2+}\) the limiting reactant, since there's an insufficient amount of \(\text{OH}^-\) to use it up completely.

Importance of limiting reactant:

• It determines the maximum amount of product that can be formed.
• After identifying it, one can evaluate the remaining excess reactants, which affects by-products and remaining solution components.

Precipitation reactions occur when ions in solution form an insoluble compound, leaving a solid precipitate. In this exercise, \(\text{Ni(OH)}_2\) is the precipitate formed when \(\text{Ni}^{2+}\) ions from \(\text{NiCl}_2\) react with \(\text{OH}^-\) ions from \(\text{NaOH}\).

The reaction \( \text{Ni}^{2+} + 2\text{OH}^- \rightarrow \text{Ni(OH)}_2 \downarrow \) shows how precipitation proceeds until the equilibrium is reached.

In a practical setting, determining when a precipitate will form and to what extent relies on calculating \(K_{sp}\) of the compound. When the ion product exceeds the \(K_{sp}\), precipitation occurs.

Significance of precipitation reactions:

• They remove ions from solution, affecting concentrations and subsequent calculations (like \(pH\)).
• Their understanding helps in predicting when and how much solid may form during a reaction process.