Question

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

Short answer

The volume of 0.100 M NaOH required to precipitate all of the nickel(II) ions from 150.0 mL of 0.249 M Ni(NO3)2 solution is 747 mL.

Step-by-step solution

Equation

The precipitate formation occurs when NaOH reacts with Ni(NO3)2 to form solid Ni(OH)2 and NaNO3. The balanced chemical equation is: \(Ni(NO_3)_2 + 2NaOH \rightarrow Ni(OH)_2 + 2NaNO_3\) Step 2: Convert the volume and concentration of Ni(NO3)2 to moles

Moles of Ni(NO3)2

Using the given volume and concentration of Ni(NO3)2, we can find the moles of Ni(NO3)2: moles = concentration × volume moles of \(Ni(NO_3)_2 = 0.249M \times 150.0mL \times \frac{1L}{1000mL}\) moles of \(Ni(NO_3)_2 = 0.03735 mol\) Step 3: Find the moles of NaOH needed

Moles of NaOH

According to the balanced chemical equation in step 1, 2 moles of NaOH is needed for every mole of Ni(NO3)2. So, we can calculate the moles of NaOH: moles of NaOH = moles of \(Ni(NO_3)_2 \times \frac{2 \, moles \, of\, NaOH}{1\, mole\, of\, Ni(NO_3)_2}\) moles of NaOH = 0.03735 mol × 2 moles of NaOH = 0.0747 mol Step 4: Calculate the volume of NaOH needed

Volume of NaOH

Now that we know the moles of NaOH needed to precipitate all the nickel(II) ions, we can find the volume of 0.100 M NaOH solution required: volume = moles / concentration volume of NaOH = 0.0747 mol / 0.100 M volume of NaOH = 0.747 L Step 5: Convert the volume of NaOH to the desired unit

Final Volume

We'll convert the volume of NaOH from liters to milliliters: volume of NaOH = 0.747 L × \(1000 \frac{mL}{L}\) volume of NaOH = 747 mL The volume of 0.100 M NaOH required to precipitate all of the nickel(II) ions from 150.0 mL of 0.249 M Ni(NO3)2 solution is 747 mL.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves using the balanced chemical equation to calculate the amount of each substance involved. A balanced equation ensures that the same number of atoms for each element is present on both sides of the equation.

In our example, we have the reaction where nickel(II) nitrate, \(\mathrm{Ni(NO_3)_2}\), reacts with sodium hydroxide, \(\mathrm{NaOH}\), to form nickel(II) hydroxide, \(\mathrm{Ni(OH)_2}\), and sodium nitrate, \(\mathrm{NaNO_3}\). The balanced chemical equation is:
\[ \mathrm{Ni(NO_3)_2 + 2NaOH \rightarrow Ni(OH)_2 + 2NaNO_3} \]

This equation tells us that one mole of nickel(II) nitrate reacts with two moles of sodium hydroxide. This is the stoichiometric ratio, and it is key in determining how much of each reactant is needed or how much product is formed.

• The coefficients in the balanced equation indicate the proportions of reactants and products.
• Stoichiometry allows us to convert quantities between moles, mass, and volume.

Precipitation Reactions

Precipitation reactions involve the formation of a solid, known as a precipitate, from the reaction of two aqueous solutions. This type of reaction occurs when the product of the reaction is insoluble in water, causing it to settle out of the solution as a solid.

In the presented chemical equation, Ni(OH)₂ is our precipitate. When the aqueous solution of nickel(II) nitrate is mixed with sodium hydroxide, nickel(II) hydroxide forms as a solid, removing nickel ions from the solution:
\[ \mathrm{Ni^{2+} + 2 OH^- \rightarrow Ni(OH)_2 (s)} \]

• Precipitation reactions are often used to isolate or purify a substance.
• These reactions are visually identifiable due to the settling of the solid out of the solution.
• The insolubility of the product in water is what drives the reaction to completion.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (mol/L). Understanding molarity is crucial for calculating how much of a reactant is needed to react completely with another in a solution.

For our exercise, we had to determine how much NaOH is required to react with a given amount of Ni(NO₃)₂. Given the molarity and volume of the nickel solution, we can find the moles of Ni(NO₃)₂:
\[ \text{Moles of Ni(NO}_3)_2 = \text{Molarity} \times \text{Volume} \]
This allows us to use stoichiometry to calculate the required moles of NaOH and, subsequently, its volume:
\[ \text{Volume of NaOH} = \frac{\text{Moles of NaOH}}{\text{Molarity of NaOH}} \]

• Molarity is a convenient way to express concentration, especially in chemical equations and reactions.
• It makes it easier to directly use stoichiometric coefficients from balanced equations.
• Always remember to convert units where needed, such as volumes from milliliters to liters.