Question

A 22.02-mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64-mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH}\). Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Short answer

Mg(NO3)2 fully reacts, NaOH excess leads to [Na+] = 0.537 M, [OH-] = 0.100 M, [NO3-] = 0.430 M.

Step-by-step solution

Write the balanced chemical equation

The reaction between magnesium nitrate \(\text{Mg(NO}_3)_2\) and sodium hydroxide \(\text{NaOH}\) is given by: \[\text{Mg(NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Mg(OH)}_2 (s) + 2\text{NaNO}_3 \]. This shows that magnesium nitrate and sodium hydroxide react to form solid magnesium hydroxide and sodium nitrate in solution.

Calculate moles of \(\text{Mg(NO}_3)_2\)

First, find the molar mass of \(\text{Mg(NO}_3)_2\), which is \(24.31 + 2(14.01 + 3\times16.00) = 148.31 \, \text{g/mol}\). The moles of \(\text{Mg(NO}_3)_2\) is calculated using the formula: \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.615 \, \text{g}}{148.31 \, \text{g/mol}} = 0.01089 \, \text{mol}\].

Calculate moles of \(\text{NaOH}\)

The molar mass of \(\text{NaOH}\) is \(22.99 + 15.99 + 1.01 = 40.00 \, \text{g/mol}\). The moles of \(\text{NaOH}\) is calculated by: \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.073 \, \text{g}}{40.00 \, \text{g/mol}} = 0.02683 \, \text{mol}\].

Determine the limiting reagent

According to the balanced equation, \(1\) mole of \(\text{Mg(NO}_3)_2\) requires \(2\) moles of \(\text{NaOH}\). Calculate moles of \(\text{NaOH}\) needed for \(0.01089\) moles of \(\text{Mg(NO}_3)_2\): \[\text{Required NaOH moles} = 2 \times 0.01089 = 0.02178^{2} \, \text{mol}\]. Since \(0.02683 > 0.02178\), \(\text{Mg(NO}_3)_2\) is the limiting reagent.

Calculate concentrations of ions after reaction

After the reaction, all \(\text{Mg(NO}_3)_2\) is consumed, so no \(\text{Mg}^{2+}\) remains. The amount of \(\text{NaOH}\) that has reacted is \(0.02178 \, \text{mol}\), leaving \(0.02683 - 0.02178 = 0.00505 \, \text{mol}\) of \(\text{NaOH}\) and thus \(0.00505\) moles of \(\text{OH}^-\) ions. The total final volume is \(50.66 \, \text{mL} = 0.05066 \, \text{L}\). Thus, the concentration of \(\text{Na}^+\) is \[\text{Concentration of Na}^+ = \frac{2 \times 0.01089 + \text{excess NaOH moles}}{\text{final volume}} = \frac{(2 \times 0.01089) + 0.00505}{0.05066} = 0.537 \, \text{M}\]. The concentration of \(\text{OH}^-\) is \[\text{Concentration of OH}^- = \frac{0.00505}{0.05066} = 0.100 \, \text{M}\]. The concentration of \(\text{NO}_3^-\) is \[\text{Concentration of NO}_3^- = \frac{2 \times 0.01089}{0.05066} = 0.430 \, \text{M}\].

Key concepts

Limiting Reagent

In chemical reactions, knowing the limiting reagent is crucial as it determines the extent of the reaction and the amount of products formed. Simply put, the limiting reagent is the reactant that gets completely consumed first, stopping the reaction from proceeding further.
For our exercise, to identify the limiting reagent, we need to compare the amount of each reactant based on the stoichiometry of the balanced chemical equation.

• Given that 1 mole of \( \text{Mg(NO}_3)_2 \) reacts with 2 moles of \( \text{NaOH} \), we must check how much \( \text{NaOH} \) is required for the available moles of \( \text{Mg(NO}_3)_2 \).
• From our calculations, 0.01089 moles of \( \text{Mg(NO}_3)_2 \) requires 0.02178 moles of \( \text{NaOH} \).

This means that the approximately 0.02683 moles of \( \text{NaOH} \) present is more than sufficient. Hence, \( \text{Mg(NO}_3)_2 \) is our limiting reagent since it is fully consumed first, limiting the reaction from continuing.

Molar Mass Calculation

Molar mass, often referred to as molecular weight, is the mass of one mole of a substance, measured in grams per mole (g/mol). This calculation is fundamental in converting between grams and moles which is a central aspect in stoichiometry.
To calculate molar mass, add up all the atomic masses of each atom present in the formula of the compound.
For example:

• For \( \text{Mg(NO}_3)_2 \), add the atomic masses: \( 24.31 \) (Mg) + \( 2 \times (14.01+3 \times 16.00) \) (NO\(_3\) groups) = 148.31 g/mol.
• Similarly, for \( \text{NaOH} \), it's \( 22.99 \) (Na) + \( 15.99 \) (O) + \( 1.01 \) (H) = 40.00 g/mol.

Understanding this concept allows for converting the given mass of a reactant into moles, which is essential for determining its proportion in a chemical reaction.

Ion Concentration

Ion concentration is how much of a particular ion is present in a solution. It's often expressed in molarity (M), which is moles of solute per liter of solution. Calculating ion concentration after a reaction helps in understanding the remaining reactants, products, and overall solution composition.
Let's calculate the ion concentrations post-reaction:

• After the reaction of \( \text{Mg(NO}_3)_2 \) and \( \text{NaOH} \), we have excess \( \text{NaOH} \), leaving \( 0.00505 \) moles of \( \text{OH}^- \) ions in solution.
• Moles of sodium ions \( \text{Na}^+ \) come from both \( \text{NaOH} \) and \( \text{NaNO}_3 \), totaling \( 0.02683 \) moles.

This gives concentrations as follows:

• \( \text{OH}^- = \frac{0.00505}{0.05066} = 0.100 \text{ M}\).
• \( \text{Na}^+ = \frac{2 \times 0.01089 + 0.00505}{0.05066} = 0.537 \text{ M}\).

Calculating these helps us assess the ionic composition of the solution.

Balanced Chemical Equation

A balanced chemical equation accurately represents the conservation of mass and charge in a chemical reaction. Each side of the equation should have equal numbers of each type of atom. Balancing is crucial as it ensures the stoichiometry—relations of moles of reactants and products—is correct for the reaction.

• For our reaction, \( \text{Mg(NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Mg(OH)}_2 (s) + 2\text{NaNO}_3 \), the equation is balanced because we maintain equal numbers of each type of atom on both sides.
• This balanced equation is used for determining stoichiometric coefficients, which tell us how many moles of each substance are involved, allowing us to identify the limiting reagent.

Understanding how to balance a chemical equation is a foundational skill for working with reactions and performing stoichiometric calculations effectively.