Question

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

Short answer

The volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to precipitate all of the nickel(II) ions from \(150.0 \mathrm{~mL}\) of a \(0.249 \mathrm{M} \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) solution is 747 mL.

Step-by-step solution

Write the balanced precipitation reaction equation

For the given problem, the precipitation reaction is: Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq)

Calculate the moles of Ni(NO₃)₂

Next, we calculate the moles of Ni(NO₃)₂ in the given solution using the concentration and volume provided: moles of Ni(NO₃)₂ = Concentration × Volume moles of Ni(NO₃)₂ = 0.249 M × 150.0 mL × \(\frac{1 L}{1000 mL}\) moles of Ni(NO₃)₂ = 0.03735 mol

Use stoichiometry to find moles of NaOH

The stoichiometry from the balanced equation indicates that for every 1 mole of Ni(NO₃)₂, 2 moles of NaOH are required. Therefore, using Ni(NO₃)₂ moles, we can find the required moles of NaOH: moles of NaOH = 2 × moles of Ni(NO₃)₂ moles of NaOH = 2 × 0.03735 mol moles of NaOH = 0.0747 mol

Calculate volume of 0.100 M NaOH solution

To find the required volume of 0.100 M NaOH solution, we can use the moles and concentration of NaOH: Volume of NaOH solution = \(\frac{moles\ of\ NaOH}{concentration\ of\ NaOH}\) Volume of NaOH solution = \(\frac{0.0747\ mol}{0.100\ M}\) Volume of NaOH solution = 0.747 L

Convert volume to milliliters

Now, we'll convert the volume of NaOH solution to milliliters: Volume of NaOH solution (mL) = 0.747 L × \(\frac{1000\ mL}{1\ L}\) Volume of NaOH solution (mL) = 747 mL So, the volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to precipitate all of the nickel(II) ions from \(150.0 \mathrm{~mL}\) of a \(0.249 \mathrm{M} \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) solution is 747 mL.

Key concepts

Precipitation Reactions

Precipitation reactions are fascinating chemical events where two soluble salts interact in a solution to form an insoluble solid known as a precipitate. This solid separates out from the solution and can often be seen settling at the bottom of the container. Such reactions are crucial in various fields such as water treatment, chemical manufacturing, and laboratory analyses.

Precipitation occurs when ions in aqueous solutions combine to form a compound that is insoluble in water. In the example exercise, nickel hydroxide (Ni(OH)₂) is the insoluble compound that precipitates out when nickel(II) nitrate (Ni(NO₃)₂) reacts with sodium hydroxide (NaOH). This equation must be balanced to ensure conservation of mass and charge.

• You'll need to identify which ions will form the solid precipitate.
• A solubility chart can help determine if the resulting compound is insoluble.

Since Ni(OH)₂ is not soluble in water, it precipitates out, illustrating a classic precipitation reaction.

Chemical Equations

Writing chemical equations is like drawing a map for a chemical reaction. It uses symbols and formulas to represent the substances involved. Balancing these equations is essential as it ensures that the number of each type of atom is the same on both sides of the equation. In our case, the balanced equation: Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq) illustrates the reaction between nickel(II) nitrate and sodium hydroxide.

For this to be balanced:

• Each side of the equation must contain the same type and number of atoms.
• The coefficients should be simplified as much as possible, while maintaining balance.

The formation of Ni(OH)₂ (s) is shown as part of the products, signifying the precipitate that forms.

Molarity

Molarity is a key concept in chemistry defined as the amount of solute divided by the volume of the solution. It is expressed in moles per liter (M). This measure allows us to depict how concentrated a solution is. For instance, having a 0.249 M solution of Ni(NO₃)₂ means there are 0.249 moles of Ni(NO₃)₂ per liter of solution. So when working with 150.0 mL, we use the equation: Moles of Ni(NO₃)₂ = Concentration × Volume in L. This allows you to find the exact number of moles of solute present in a given volume of solution, which is crucial for stoichiometric calculations.

Solution Volume Calculation

Calculating solution volume is often necessary to solve chemical equations and reactions. It involves using the moles of solute and the desired molarity to determine the volume needed of a solution to fully react with another substance.

In our example, we converted the moles of NaOH required for the reaction to find the volume. By using the formula: Volume (L) = \(\frac{moles}{concentration}\)we determined how much NaOH solution in liters is required. This was then converted to milliliters for practical measurements, since volumes are often more manageable in smaller units during laboratory work.

• Always keep track of the units involved, converting where necessary to keep consistency.
• Such calculations are fundamental in ensuring sufficient quantities are used for a reaction to proceed completely.