Question

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Short answer

(a) The balanced chemical equation is: \(2 \mathrm{KOH} + \mathrm{NiSO}_{4} \longrightarrow \mathrm{Ni}( \mathrm{OH})_{2} + \mathrm{K}_{2} \mathrm{SO}_{4}\). (b) The precipitate formed is Nickel(II) hydroxide, Ni(OH)₂. (c) KOH is the limiting reactant. (d) The mass of Ni(OH)₂ precipitate formed is 0.9271 g. (e) The concentrations of remaining ions in the solution are: Ni²⁺: 0.0667 M, K⁺: 0.1333 M, and SO₄²⁻: 0.0667 M.

Step-by-step solution

Write the balanced chemical reaction

The balanced chemical reaction between KOH and NiSO4 is as follows: \(2 \mathrm{KOH} + \mathrm{NiSO}_{4} \longrightarrow \mathrm{Ni}( \mathrm{OH})_{2} + \mathrm{K}_{2} \mathrm{SO}_{4}\)

Identify the precipitate

The precipitate formed will be the insoluble product in the balanced chemical reaction, which is Nickel(II) hydroxide, Ni(OH)₂.

Determine the Limiting Reactant

Calculate the initial moles of each reactant: Moles of KOH in the solution: \(0.200 \mathrm{M} × 0.100 \mathrm{L} =0.020 \mathrm{mol}\) Moles of NiSO4 in the solution: \(0.150 \mathrm{M} × 0.200 \mathrm{L} = 0.030 \mathrm{mol}\) The stoichiometric ratio of KOH to NiSO₄ is 2:1. So, divide the moles of each reactant by their respective coefficients: KOH: \(\frac{0.020}{2} = 0.010\) NiSO₄: \(\frac{0.030}{1} = 0.030\) Since the smallest ratio is associated with KOH, it is the limiting reactant.

Calculate the mass of the precipitate

To determine the number of moles of Ni(OH)₂ formed, use stoichiometric ratios: \(\mathrm{Mole \ ratio= \frac{Moles \ of \ Ni(OH)_{2}}{Moles \ of \ KOH} = \frac{1}{2}}\) Moles of Ni(OH)₂ formed: \(0.020 \mathrm{mol \ KOH} × \frac{1 \ mol \ Ni(OH)_{2}}{2 \ mol \ KOH} = 0.010 \mathrm{mol\ Ni(OH)_{2}}\) Now, calculate the mass of Ni(OH)₂ formed using the molar mass of Ni(OH)₂ (92.71 g/mol): Mass of Ni(OH)₂ formed: \(0.010 \mathrm{mol\ Ni(OH)_{2}} × 92.71 \frac{\mathrm{g}}{\mathrm{mol}} = 0.9271 \mathrm{g}\)

Calculate the concentration of ions remaining in the solution

First, let's determine the amount of NiSO₄ remaining after the reaction: Moles of NiSO₄ remaining: \(0.030 \mathrm{mol} - 0.010 \mathrm{mol} = 0.020 \mathrm{mol}\) Now, calculate the molality of the excess reactant, Ni²⁺, and the products, K⁺ and SO₄²⁻: Ni²⁺: \(\frac{0.020 \ \mathrm{mol}}{0.100 + 0.200} \ \mathrm{L} = 0.0667\ \mathrm{M}\) K⁺: All moles of KOH have reacted, forming 2 moles of K⁺ from each mol of KOH. \(\frac{2 \times 0.020 \ \mathrm{mol}}{0.100 + 0.200} \ \mathrm{L} = 0.1333\ \mathrm{M}\) SO₄²⁻: \(\frac{0.020 \ \mathrm{mol}}{0.100 + 0.200} \ \mathrm{L} = 0.0667\ \mathrm{M}\) The concentrations of remaining ions in the solution are: Ni²⁺: 0.0667 M, K⁺: 0.1333 M, and SO₄²⁻: 0.0667 M.

Key concepts

Limiting Reactant Calculation

Understanding the limiting reactant in a chemical reaction is essential for predicting the amount of product that will form. The limiting reactant is the reactant that is completely consumed first during the reaction, preventing further product from being created. In the provided exercise, we've calculated it by comparing the mole ratio of the reactants to the ratio in the balanced equation.

In our case, KOH and NiSO4 react in a 2:1 stoichiometric ratio, which means two moles of KOH are required for every mole of NiSO4 to react completely. By dividing the initial moles of KOH by 2, and NiSO4 by 1, the lowest result shows which reactant will limit the reaction. KOH is identified as the limiting reactant because it has the smallest available mole ratio. This step is crucial to determine how much Ni(OH)2, the precipitate, will be formed.

This process helps in minimizing waste and optimizing the use of reactants in industrial applications. When attempting to solve problems involving limiting reactants, it's always important to start by writing the balanced chemical equation and then proceed to calculate the moles of each reactant present.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form at least one insoluble product, or precipitate. This is useful both in industrial processes and analytical chemistry where precipitation is often used to isolate or remove specific elements or compounds.

In the referenced exercise, when KOH reacts with NiSO4, the insoluble product formed is Ni(OH)2. This is identified through knowledge of solubility rules or consulting a solubility chart, which tells us that hydroxides are generally insoluble, with some exceptions. During a precipitation reaction, ions come together to form a solid that falls out of the solution. The solid can be filtered out, and its mass can be used to determine how much of the substance was present originally.

To further improve understanding, drawing a molecular diagram showing the ions in solution before and after the reaction or even performing a hands-on experiment in a safe lab setting can make this concept clearer.

Molar Mass Calculation

The molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance (usually in grams per mole). It is calculated by adding the average atomic masses of the constituent elements found in the periodic table, taking into account their respective proportions in the molecule.

For instance, to calculate the mass of the precipitate, Ni(OH)2, we use its molar mass. This is done by multiplying the atomic mass of nickel by one (since there's one nickel atom per formula unit), the atomic mass of oxygen by two, and the atomic mass of hydrogen by two. The molar mass is pivotal in converting between the mass of a substance and the number of moles, which is a standard step in stoichiometric calculations. In our exercise, after finding the number of moles of Ni(OH)2, we multiplied it by the molar mass to find the mass in grams.

Grasping the concept of molar mass is important as it lays the foundation for understanding more complex topics such as reaction yields, molecular weight distributions, and conversion of units within chemical equations.