Problem 59 - Chemistry and Chemical Reactivity

Question

Stoichiometry of Reactions in Solution What volume of \(0.109 \mathrm{M} \mathrm{HNO}_{3},\) in milliliters, is required to react completely with \(2.50 \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\) \(2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\) $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) $$

Step-by-Step Solution

Answer

268 mL of 0.109 M HNO3 is required.

Step 1: Calculate Moles of Ba(OH)2

First, determine the moles of Ba(OH)2 using its molar mass. The molar mass of Ba(OH)2 is approximately 171.34 g/mol. Use the formula: \[ n = \frac{\text{mass}}{\text{molar mass}} \] Given that the mass is 2.50 g, the moles of Ba(OH)2 are calculated as:\[ n = \frac{2.50 \, \text{g}}{171.34 \, \text{g/mol}} = 0.0146 \, \text{mol} \]

Step 2: Use Stoichiometry to Find Moles of HNO3

From the balanced chemical equation, it is clear that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3. Use this stoichiometric relationship to find the moles of HNO3 needed:\[ n_{\text{HNO}_3} = 2 \times n_{\text{Ba(OH)}_2} = 2 \times 0.0146 \, \text{mol} = 0.0292 \, \text{mol} \]

Step 3: Calculate Volume of HNO3 Solution

Use the concentration equation to find the volume of the HNO3 solution. Recall that \[ C = \frac{n}{V} \] , where \(C\) is the concentration in molarity (M), \(n\) is the number of moles, and \(V\) is the volume in liters. Rearrange it to find volume:\[ V = \frac{n}{C} = \frac{0.0292 \text{ mol}}{0.109 \text{ M}} = 0.268 \, \text{L} \]

Step 4: Convert Volume to Milliliters

Finally, convert the volume from liters to milliliters by multiplying by 1000:\[ V = 0.268 \, \text{L} \times 1000 = 268 \, \text{mL} \]

Key Concepts

Reactions in Solution

Reactions in solution are a fundamental aspect of chemistry. They occur when reactants dissolve in a solvent to form a solution, typically involving liquid or aqueous phases. Understanding these reactions helps us predict the result of mixing different substances together. When solving problems involving reactions in solutions, it's crucial to consider:

• The identity and state (solid, liquid, gas, aqueous) of each reactant and product as demonstrated by the given equation.
• Whether the reaction happens through dissolution, where compounds are broken down into ions in water.
• The role of the solvent, which is usually water in aqueous solutions, facilitating the interaction of reactants to form products.

In our exercise, nitric acid \(\text{HNO}_3\) reacts with solid barium hydroxide \(\text{Ba(OH)}_2\), both typically dissolved in water. The balanced equation provided is crucial as it shows the stoichiometry necessary to solve various calculations, such as determining the required reactant volume.

Molar Mass Calculation

Calculating molar mass is an essential step in many stoichiometry problems. Molar mass refers to the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). To find the molar mass of a compound, sum up the atomic masses of all the atoms present in its formula, based on the periodic table values.For example, to calculate the molar mass of barium hydroxide \(\text{Ba(OH)}_2\):

• Barium (Ba): Has an atomic mass of approximately 137.33 g/mol.
• Oxygen (O): Has an atomic mass of about 16.00 g/mol.
• Hydrogen (H): Has an atomic mass around 1.01 g/mol.

Combine these based on the formula \(\text{Ba(OH)}_2\):\[\text{Molar mass of} \ \text{Ba(OH)}_2 = 137.33 + (2 \times (16.00 + 1.01)) = 171.34 \, \text{g/mol}\]This calculation is pivotal in determining the number of moles from a given mass, which is a key step in stoichiometric calculations.

Chemical Equation Balancing

Balancing chemical equations is a critical skill in chemistry. It ensures that the number of atoms for each element is equal on both sides of the reaction, maintaining the law of conservation of mass. To balance an equation:

• Identify reactants and products, writing their correct formulas.
• Adjust coefficients (numbers in front of molecules) to have the same number of atoms of each element on both sides.
• Start by balancing elements that appear in only one reactant and one product.
• Balance complex molecules and leave solitary elements, like O or H, for last.

In the given exercise, \(2 \, \text{HNO}_3(\text{aq}) + \text{Ba(OH)}_2(\text{s}) \rightarrow 2 \, \text{H}_2\text{O}(\ell) + \text{Ba(NO}_3\text{)}_2(\text{aq})\), each side corresponds atomically, maintaining the balance:- 2 moles of \(\text{HNO}_3\) for each mole of \(\text{Ba(OH)}_2\).The visually balanced equation helps us in the following stoichiometry steps.

Titration Calculations

Titration is an analytical method used to determine the concentration of a solute in a solution. During titration, a solution of known concentration is used to react with a solution of unknown concentration, enabling us to find its concentration.In our example, although it's not a conventional titration setup, we can think of the reaction between \(\text{HNO}_3\) and \(\text{Ba(OH)}_2\) as similar. We're essentially calculating the needed volume of one reactant (acid solution) to completely react with a given amount of another reactant (base).Key principles include:

• The equivalence point: When moles of acid equal moles of the base, as per the balanced chemical equation.
• Using the molarity formula \(C = \frac{n}{V}\) to calculate unknowns. Rearrange it for volume \(V\) when moles \(n\) and concentration \(C\) are known, as done in the steps.
• Conversions, such as converting liters to milliliters after calculations are complete.

In the exercise, calculated moles and molarity allow us to determine that 268 mL of \(0.109 \, M\) \(\text{HNO}_3\) is required to react completely with the barium hydroxide.