Question

What volume of 0.100 \(M\) NaOH is required to precipitate all of the nickel(II) ions from \(150.0 \mathrm{mL}\) of a \(0.249-M\) solution of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short answer

747.0 mL of 0.100 M NaOH solution is required to precipitate all the nickel(II) ions from 150.0 mL of a 0.249 M Ni(NO₃)₂ solution.

Step-by-step solution

Write the balanced chemical equation for the reaction.

We want to know the reaction between NaOH and Ni(NO₃)₂: Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq)

Find the number of moles of Ni(NO₃)₂.

Given that we have 150.0 mL of 0.249 M Ni(NO₃)₂, we can find the moles of \(Ni(NO_3)_2\). First, we need to convert the volume to liters, then multiply by the provided molarity: Moles of \(Ni(NO_3)_2\) = volume (in L) × molarity Moles of \(Ni(NO_3)_2\) = 0.150 L × 0.249 mol/L = 0.03735 mol

Calculate the moles of NaOH needed

Using the stoichiometry from the balanced chemical equation (1 Ni(NO₃)₂ requires 2 NaOH), we can find the number of moles of NaOH needed: Moles of NaOH = moles of \(Ni(NO_3)_2\) × 2 moles of NaOH / 1 mole of \(Ni(NO_3)_2\) Moles of NaOH = 0.03735 mol × 2 = 0.07470 mol

Determine the volume of NaOH needed

Now that we have the moles of NaOH needed, we can use the given molarity of NaOH to find the required volume (in L): Volume of NaOH = moles of NaOH ÷ molarity of NaOH Volume of NaOH = 0.07470 mol ÷ 0.1 M = 0.747 L Since we want the answer in mL, we'll multiply by 1000: Volume of NaOH = 0.747 L × 1000 = 747.0 mL Hence, 747.0 mL of the 0.100 M NaOH solution would be required to precipitate all the nickel(II) ions from 150.0 mL of the 0.249 M Ni(NO₃)₂ solution.

Key concepts

Balanced Chemical Equation

In any chemical reaction, it is essential to write a balanced chemical equation to understand how reactants transform into products. Balancing chemical equations ensures that the number of atoms of each element is the same on both sides of the equation. This follows the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction.

For our problem, the balanced chemical equation is:

• Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq)

This equation tells us that one molecule of nickel(II) nitrate reacts with two molecules of sodium hydroxide to produce one molecule of nickel(II) hydroxide and two molecules of sodium nitrate. Observing such equations helps us determine the ratios in which reactants combine and products form, which is crucial for stoichiometry.

Moles Calculation

Calculating moles is a fundamental step in solving stoichiometry problems. Moles provide a bridge between the atomic world and the macroscopic world we observe. To convert volume and molarity to moles, use the formula:

\[\text{Moles} = \text{Volume (in L)} \times \text{Molarity (mol/L)}\]For instance, when calculating the moles of Ni(NO₃)₂:

• Convert volume from milliliters to liters: 150.0 mL = 0.150 L
• Multiply by the molarity: 0.150 L × 0.249 mol/L = 0.03735 mol.

This step ensures you know how many moles of a substance are present, which you can then use to determine the amount of another reactant or product involved in the chemical reaction.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions react to form an insoluble solid, known as the precipitate. This reaction is a type of double displacement reaction. In this specific exercise, when sodium hydroxide (NaOH) reacts with nickel(II) nitrate (Ni(NO₃)₂), nickel(II) hydroxide (Ni(OH)₂) precipitates out of the solution:

• Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq)

The solid nickel(II) hydroxide forms as a result of the aqueous nickel(II) ions and hydroxide ions coming together. Precipitation is a useful process in chemistry to remove a specific ion from a solution, such as in this exercise to remove nickel(II) ions.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution and is expressed as mol/L or simply M. Understanding molarity allows you to determine how strong or dilute a solution is. The formula for calculating molarity is:

\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]In the problem at hand, we are dealing with a 0.100 M NaOH solution. This means there are 0.100 moles of NaOH in every liter of solution. By applying molarity in stoichiometry calculations, as seen in our exercise, you can find out how much of a reagent is required to complete a reaction. For example, knowing the moles of NaOH needed, we calculate the required volume of 0.100 M NaOH solution to achieve the desired reaction outcome.