Problem 143 - Chemistry: Structure and Properties

Question

What volume of 0.100 M sodium carbonate solution is required to precipitate 99% of the Mg from 1.00 L of 0.100 M magnesium nitrate solution?

Step-by-Step Solution

Answer

To precipitate 99% of the Mg from 1.00 L of 0.100 M magnesium nitrate solution, 0.99 L of 0.100 M sodium carbonate solution is required.

Step 1: Write down the balanced chemical equation

Identify the reactants and products and balance the chemical equation. The reaction between magnesium nitrate and sodium carbonate to produce magnesium carbonate and sodium nitrate is: \[ \text{Mg(NO_3)_2} + \text{Na}_2\text{CO}_3 \rightarrow \text{MgCO}_3 + 2 \text{NaNO}_3 \]

Step 2: Determine the moles of magnesium present

Calculate the moles of magnesium nitrate using the concentration and volume of the solution. The molar mass of magnesium nitrate is not needed as the reaction is in a 1:1 ratio. Using the formula \( n = M \times V \), where \( n \) is the number of moles, \( M \) is molarity, and \( V \) is volume in liters, we get \[ n_{\text{Mg(NO_3)_2}} = 0.100 \text{ M} \times 1.00 \text{ L} = 0.100 \text{ mol} \]

Step 3: Calculate the moles of magnesium to be precipitated

Determine how many moles of magnesium will be precipitated (99% of the initial moles). \[ n_{\text{precipitated Mg}} = 0.99 \times n_{\text{Mg(NO_3)_2}} = 0.99 \times 0.100 \text{ mol} = 0.099 \text{ mol} \]

Step 4: Calculate moles of sodium carbonate required

From the balanced equation, 1 mole of Mg(NO3)2 reacts with 1 mole of Na2CO3. Thus, the moles of Na2CO3 needed is the same as the moles of precipitated Mg. \[ n_{\text{Na}_2\text{CO}_3} = n_{\text{precipitated Mg}} = 0.099 \text{ mol} \]

Step 5: Calculate the volume of sodium carbonate solution

Utilize the molarity formula to find the volume of the Na2CO3 solution required. \[ V_{\text{Na}_2\text{CO}_3} = \frac{n_{\text{Na}_2\text{CO}_3}}{M_{\text{Na}_2\text{CO}_3}} = \frac{0.099 \text{ mol}}{0.100 \text{ M}} = 0.99 \text{ L} \]

Key Concepts

Balancing Chemical Equations

Understanding how to balance a chemical equation is a fundamental skill in chemistry. A balanced chemical equation ensures that the same number of atoms of each element is present on both the reactant and product sides of the equation. This reflects the Law of Conservation of Mass, which states that matter cannot be created or destroyed in an ordinary chemical reaction.

To balance an equation, first, write down the unbalanced chemical equation. Then, adjust the stoichiometric coefficients — the numbers in front of each reactant and product — to achieve an equal number of each type of atom on both sides of the arrow. For example, the reaction between magnesium nitrate and sodium carbonate, producing magnesium carbonate and sodium nitrate, is initially unbalanced. With careful adjustment, we establish the balanced equation:
\[ \text{Mg(NO_3)_2} + \text{Na}_2\text{CO}_3 \rightarrow \text{MgCO}_3 + 2 \text{NaNO}_3 \]
Here, one molecule of magnesium nitrate reacts with one molecule of sodium carbonate to produce one molecule of magnesium carbonate and two molecules of sodium nitrate, indicating a 1:1:1:2 molar ratio.

Molarity Calculations

Molarity, symbolized by 'M', is a measure of concentration in chemistry that indicates the number of moles of solute per liter of solution. It is a crucial concept for quantitative analysis in solutions.

To calculate molarity, you can use the formula:
\[ M = \frac{n}{V} \]
where \( M \) represents molarity, \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters. In the given problem, the molarity of magnesium nitrate solution is provided as 0.100 M. To find the moles of contained magnesium nitrate, we apply the molarity formula and discover that 1.00 L of solution contains 0.100 moles of the solute:
\[ n_{\text{Mg(NO_3)_2}} = 0.100 \text{ M} \times 1.00 \text{ L} = 0.100 \text{ mol} \]

Precipitation Reactions

A precipitation reaction is a type of chemical reaction where a solid, called a precipitate, forms from the mixture of two solutions. The precipitate is usually an insoluble ionic compound. When two aqueous solutions containing soluble salts are mixed, the formation of an insoluble salt can occur if the product of the ions' concentrations exceeds the solubility product (Ksp) of the compound.

In the exercise, when a solution of sodium carbonate is added to magnesium nitrate, magnesium carbonate, which is a sparingly soluble compound, precipitates out, leaving sodium nitrate in solution. The completion of precipitation is driven by the stoichiometry of the reaction and is quantitatively described by the percentage of the reactant converted into the precipitate. In our problem, we aim to precipitate 99% of magnesium from magnesium nitrate, which we calculate as follows:
\[ n_{\text{precipitated Mg}} = 0.99 \times n_{\text{Mg(NO_3)_2}} = 0.99 \times 0.100 \text{ mol} = 0.099 \text{ mol} \]

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers used to balance a chemical equation, signifying the proportional amounts of reactants and products involved in the reaction. These coefficients allow us to convert between moles of one substance and moles of another, based on the balanced chemical equation.

For instance, the stoichiometry of the reaction between sodium carbonate and magnesium nitrate is 1:1, indicating that one mole of sodium carbonate reacts with one mole of magnesium nitrate to yield products. By knowing the stoichiometry, we can determine the amount of sodium carbonate needed to precipitate a certain amount of magnesium. Here's how we calculate it for our example: Since 0.099 mol of magnesium carbonates to form, and the stoichiometry is 1:1, we need 0.099 mol of sodium carbonate:
\[ n_{\text{Na}_2\text{CO}_3} = n_{\text{precipitated Mg}} = 0.099 \text{ mol} \]. This is essential for converting the moles of reactants to the required volume of reactant solution, leading us to solve the problem with the given molarity.