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. \(\left(\mathrm{K}_{\text {s }}\right.\) of \(\mathrm{Ni}(\mathrm{OH})_{2}=1.6 \times 10^{-14}\) ).

Step-by-Step Solution

Answer

The approximate pH of the final solution is 12.94.

Step 1: Calculate moles of NaOH

First, find the number of moles of NaOH. Given that the molecular weight of NaOH is 40 g/mol, calculate the moles:\[\text{Moles of NaOH} = \frac{1.75 \text{ g}}{40 \text{ g/mol}} = 0.04375 \text{ mol}\]

Step 2: Calculate moles of NiCl2

Determine the number of moles of NiCl2 in the 250 ml of the 0.1 M solution. Use the formula for moles: \[\text{Moles of NiCl}_2 = 0.1 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ mol}\]

Step 3: Determine the limiting reactant

To find the limiting reactant, the chemical reaction must be considered: \[\text{NiCl}_2 + 2 \text{NaOH} \rightarrow \text{Ni(OH)}_2 + 2 \text{NaCl}\]From the stoichiometry, 2 moles of NaOH react with 1 mole of NiCl2. Calculate moles needed:\[2 \times 0.025 \text{ mol} = 0.050 \text{ mol} \text{ of NaOH needed}\]Since only 0.04375 moles of NaOH are available, NaOH is the limiting reactant.

Step 4: Calculate excess nickel ion (Ni2+)

After the reaction, calculate the remaining amount of NiCl2:\[\text{Moles of NiCl}_2 \text{ left} = 0.025 \text{ mol} - \frac{0.04375}{2} \text{ mol} = 0.025 \text{ mol} - 0.021875 \text{ mol} = 0.003125 \text{ mol}\]

Step 5: Determine hydroxide ion concentration

The hydroxide ions (OH-) from unreacted NaOH determine the pH. Calculate moles of OH- left unreacted (since it exceeds the Ni2+ requirement):\[\text{Excess OH}^- = 0.04375 - 0.021875 = 0.021875 \text{ mol}\]Given that the volume of the solution is now approximately 250 ml (or 0.250 L), theconcentration of OH- is:\[[\text{OH}^-] = \frac{0.021875 \text{ mol}}{0.250 \text{ L}} = 0.0875 \text{ M}\]

Step 6: Calculate pOH and then pH

Convert the concentration of hydroxide ions into pOH:\[\text{pOH} = -\log_{10}(0.0875)\approx 1.06\]Finally, calculate the pH:\[\text{pH} = 14 - \text{pOH} = 14 - 1.06 = 12.94\]

Key Concepts

Limiting Reactant

In chemical reactions, it is essential to determine which reactant will be completely consumed first, as this dictates the extent of the reaction. This reactant is known as the limiting reactant. In the reaction between NaOH and NiCl₂ to form Ni(OH)₂, we compare the mole ratios.

• From the balanced equation: \[ ext{NiCl}_2 + 2 ext{NaOH} \rightarrow ext{Ni(OH)}_2 + 2 ext{NaCl}\]
• We see that 2 moles of NaOH are needed to react with 1 mole of NiCl₂.
• Given that there are 0.04375 moles of NaOH and 0.025 moles of NiCl₂ initially available, NaOH will determine the completion of the reaction.

NaOH is the limiting reactant because only 0.04375 moles are present, which is less than the needed 0.050 moles based on the stoichiometry. Understanding the limiting reactant helps predict how much product will form.

Moles of NaOH

To solve any chemical reaction problem involving quantities, we need to determine the number of moles of each reactant. Knowing the moles allows us to analyze how much of each substance is needed or remains after the reaction.

Calculating Moles

The number of moles can be calculated using the formula:\[\text{Moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}.\]In this problem, NaOH has a mass of 1.75 g and a molar mass of 40 g/mol. Plug these values into the formula:\[\text{Moles of NaOH} = \frac{1.75 \text{ g}}{40 \text{ g/mol}} = 0.04375 \text{ mol}\]Knowing the moles is crucial for predicting how reactants interact in a chemical equation.

Moles of NiCl₂

Just like NaOH, we also need to calculate the moles of NiCl₂. This step is vital to determine how much NiCl₂ can participate in the reaction.

Solution to Moles

NiCl₂ is provided in a solution, and its concentration is given in molarity (M), which is moles per liter. To find moles:\[\text{Moles of NiCl}_2 = \text{Molarity} \times \text{Volume in Liters}\]For a 0.1 M NiCl₂ solution, with a volume of 250 ml (or 0.250 L):\[\text{Moles of NiCl}_2 = 0.1 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ mol}\]Now that we know the moles of both reactants, we can easily determine which one is the limiting reactant.

Hydroxide Ion Concentration

After a chemical reaction, hydroxide ions (OH⁻) may remain in solution as an excess from the limiting reactant. These ions significantly affect the solution's pH.

Calculating Hydroxide Ion Concentration

Once it is known that NaOH is the limiting reactant, we must find out how many moles ofOH⁻ result from it. After accounting for the use of moles to fully react with NiCl₂, there is some amount of NaOH left, which contributes extraOH⁻ions:\[\text{Excess } OH^- = 0.04375 \text{ moles of NaOH} - 0.021875 \text{ moles (reacted)} = 0.021875 \text{ mol}\]The concentration of these ions is obtained by dividing the excess moles by the total volume (0.250 L):\[[OH^-] = \frac{0.021875 \text{ mol}}{0.250 \text{ L}} = 0.0875 \text{ M}\]With this concentration, we can calculate pOH, and eventually, the pH of the solution, revealing the basic environment due to leftover hydroxide ions.