Problem 245 - A Complete Resource Book in Chemistry for JEE Main

Question

\(1.75 \mathrm{gm}\) of solid \(\mathrm{NaOH}\) are added to \(250 \mathrm{ml}\) of \(0.1 \mathrm{M}\) \(\mathrm{NiCl}_{2}\) solution. Calculate the approximate \(\mathrm{pH}\) of final solution. \(\left(\mathrm{K}_{\mathrm{sp}}\right.\) of \(\left.\mathrm{Ni}(\mathrm{OH})_{2}=1.6 \times 10^{-14}\right)\).

Step-by-Step Solution

Answer

The final pH of the solution is approximately 7.

Step 1: Calculating Moles of NaOH

First, we convert grams of NaOH to moles. Use the molecular weight of NaOH, which is approximately 40 g/mol.\[\text{Moles of NaOH} = \frac{1.75 \text{ g}}{40 \text{ g/mol}} = 0.04375 \text{ moles}\]

Step 2: Calculating Moles of NiCl3

Calculate the moles of \(\text{NiCl}_2\) in 250 mL of 0.1 M solution.\[\text{Moles of NiCl}_2 = 0.1 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ moles}\]

Step 3: Reaction Between NaOH and NiCl3

\(\text{NaOH}\) reacts with \(\text{NiCl}_2\) to form \(\text{Ni(OH)}_2\), which precipitates, and \(\text{NaCl}\) remains in solution. The balanced chemical equation is:\[\text{Ni}^{2+} + 2\text{OH}^- \rightarrow \text{Ni(OH)}_2 \downarrow\]Initially, there are 0.025 moles of \(\text{Ni}^{2+}\) available to react with 0.04375 moles of \(\text{OH}^-\). Since \(\text{Ni}^{2+}\) requires 2 moles of \(\text{OH}^-\) for complete reaction, the limiting reactant is \(\text{Ni}^{2+}\).

Step 4: Determining Remaining Hydroxide Ions

After 0.025 moles of \(\text{Ni}^{2+}\) react, \[ 0.025 \times 2 = 0.05 \text{ moles of } \text{OH}^- \]are required to completely react, but only 0.04375 moles were initially available. Thus, all \(\text{Ni}^{2+}\) is consumed, leaving \[0.04375 - 0.05 = 0 \]\(\text{OH}^-\). Since all \(\text{Ni}^{2+}\) is reacted, no additional OH- ions are left free in solution due to \(\text{Ni(OH)}_2\) precipitating.

Step 5: Solution Concentration

After NaOH reacts, the final solution mainly consists of NaCl and potentially some marginally dissolved \(\text{Ni(OH)}_2\). The exact \(\text{OH}^-\) concentration can be negligible compared to the excess initially added, as they instigate precipitation. Thus, approximate pH can be based on equilibrium of \(\text{Ni(OH)}_2\) alone.

Step 6: Calculating pH of the Solution

\(\text{Ni(OH)}_2\) is sparingly soluble, with \[\text{K}_{\text{sp}} = [\text{Ni}^{2+}][\text{OH}^-]^2\]Using \[ \text{K}_{\text{sp}}=1.6 \times 10^{-14} \] We have \[ [\text{OH}^-]^2 = \frac{1.6 \times 10^{-14}}{\text{C}[\text{Ni}^{2+}]} \]Without sure C, assume significant \(\text{OH}^-\) neutralization implies low hydroxide presence for pH calculation with negligible impact, suggesting \[[\text{OH}^-] \approx 10^{-7} \text{M}, \] thus, \(pH \approx 7\) considering dilution effects and precipitation loss.

Key Concepts

Molarity

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

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.

Limiting Reactant

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 Reaction

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.