Problem 82 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of a \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what is the molarity of the \(\mathrm{BaCl}_{2}\) solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{MHCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

Step-by-Step Solution

Answer

(a) \(83.75 \mathrm{~mL}\) of 0.120 M HCl are needed to neutralize the 50.0 mL of 0.101 M Ba(OH)₂ solution. (b) \(20.0 \mathrm{~mL}\) of 0.125 M H₂SO₄ are needed to neutralize 0.200 g of NaOH. (c) The molarity of the BaCl₂ solution is \(0.0950 \mathrm{M}\). (d) The mass of Ca(OH)₂ in the solution is \(0.328 \mathrm{~g}\).

Step 1: Balance the chemical equation

The balanced chemical equation for the neutralization reaction between HCl and Ba(OH)2 is: \[2 \mathrm{HCl} + \mathrm{Ba(OH)_2} \rightarrow \mathrm{BaCl_2} + 2 \mathrm{H_2O}\]

Step 2: Calculate the moles of Ba(OH)2

Using the given volume and concentration of Ba(OH)2, we can find the moles of Ba(OH)2: Moles of Ba(OH)2 = volume × molarity Moles of Ba(OH)2 = 50.0 mL × 0.101 M = 0.00505 mol

Step 3: Find the moles of HCl needed

From the balanced equation, we know that 2 moles of HCl react with 1 mole of Ba(OH)2. Therefore, moles of HCl needed = 2 × moles of Ba(OH)2 = 2 × 0.00505 mol = 0.0101 mol

Step 4: Calculate the volume of HCl required

Using the moles of HCl and its molarity, we can find the volume of HCl needed: Volume of HCl = moles of HCl / molarity Volume of HCl = 0.0101 mol / 0.120 M = \(83.75 \mathrm{~mL}\) (b)

Step 1: Balance the chemical equation

The balanced chemical equation for the neutralization reaction between H2SO4 and NaOH is: \[2 \mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2 \mathrm{H_2O}\]

Step 2: Calculate the moles of NaOH

Using the given mass and molar mass of NaOH (40.00 g/mol), we can find the moles of NaOH: Moles of NaOH = mass / molar mass Moles of NaOH = 0.200 g / 40.00 g/mol = 0.00500 mol

Step 3: Find the moles of H2SO4 needed

From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, moles of H2SO4 needed = moles of NaOH / 2 = 0.00500 mol / 2 = 0.00250 mol

Step 4: Calculate the volume of H2SO4 required

Using the moles of H2SO4 and its molarity, we can find the volume of H2SO4 needed: Volume of H2SO4 = moles of H2SO4 / molarity Volume of H2SO4 = 0.00250 mol / 0.125 M = \(20.0 \mathrm{~mL}\) (c)

Step 1: Balance the chemical equation

The balanced chemical equation for the precipitation reaction between Na2SO4 and BaCl2 is: \[\mathrm{Na}_2\mathrm{SO}_4 + \mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 \downarrow+ 2 \mathrm{NaCl}\]

Step 2: Calculate the moles of Na2SO4

Using the given mass and molar mass of Na2SO4 (142.04 g/mol), we can find the moles of Na2SO4: Moles of Na2SO4 = mass / molar mass Moles of Na2SO4 = 752 mg / 142.04 g/mol = 0.00530 mol (note: convert from mg to g)

Step 3: Find the moles of BaCl2 needed

From the balanced equation, we know that 1 mole of BaCl2 reacts with 1 mole of Na2SO4. Therefore, moles of BaCl2 needed = moles of Na2SO4 = 0.00530 mol

Step 4: Calculate the molarity of BaCl2

Using the moles of BaCl2 and the given volume, we can find the molarity of BaCl2: Molarity of BaCl2 = moles of BaCl2 / volume Molarity of BaCl2 = 0.00530 mol / 55.8 mL = \(0.0950 \mathrm{M}\) (note: convert from mL to L) (d)

Step 1: Balance the chemical equation

The balanced chemical equation for the neutralization reaction between HCl and Ca(OH)2 is: \[2 \mathrm{HCl} + \mathrm{Ca(OH)_2} \rightarrow \mathrm{CaCl_2} + 2 \mathrm{H_2O}\]

Step 2: Calculate the moles of HCl

Using the given volume and concentration of HCl, we can find the moles of HCl: Moles of HCl = volume × molarity Moles of HCl = 42.7 mL × 0.208 M = 0.00888 mol

Step 3: Find the moles of Ca(OH)2 needed

From the balanced equation, we know that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, moles of Ca(OH)2 needed = moles of HCl / 2 = 0.00888 mol / 2 = 0.00444 mol

Step 4: Calculate the mass of Ca(OH)2

Using the moles of Ca(OH)2 and its molar mass (74.09 g/mol), we can find the mass of Ca(OH)2: Mass of Ca(OH)2 = moles of Ca(OH)2 × molar mass Mass of Ca(OH)2 = 0.00444 mol × 74.09 g/mol = \(0.328 \mathrm{~g}\)

Key Concepts

Neutralization Reactions

Neutralization reactions are a fundamental concept in chemistry. They occur when an acid reacts with a base to produce a salt and water. These reactions are crucial in various applications ranging from titrations in labs to processes in everyday life. In simple terms, neutralization is the process where acidic nature is cancelled out by the basic nature of a substance, resulting in a neutral solution. Consider the reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)₂): \[2 \mathrm{HCl} + \mathrm{Ba(OH)_2} \rightarrow \mathrm{BaCl_2} + 2 \mathrm{H_2O} \] The hydrogen ions from HCl combine with the hydroxide ions from Ba(OH)₂ to form water, while the remaining ions form barium chloride. Each acid-base pair has its unique ratio, which is critical to achieving complete neutralization. Being aware of the stoichiometry of such reactions helps in predicting the outcome of mixing particular volumes and concentrations of acids and bases.

Molarity Calculations

Molarity (\( M \)) is defined as the number of moles of solute per liter of solution. Molarity calculations are essential for determining how many moles of a reactant are needed, or how many are present in a given volume of solution, especially in applications like neutralization reactions. The basic formula for molarity is: - \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \) To demonstrate, consider calculating the volume of HCl needed to neutralize Ba(OH)₂. - First, determine the moles of Ba(OH)₂ using its volume and molarity.- Use the balanced chemical equation to find out how many moles of HCl are required.- Finally, calculate the volume of HCl by rearranging the molarity equation: \( \text{Volume of HCl} = \frac{\text{moles of HCl}}{\text{molarity of HCl}} \)This concept is repetitively used in chemistry, especially in titrations and preparing solutions.

Balanced Chemical Equations

Balanced chemical equations are vital because they represent the actual chemical reactions occurring. Each molecule or atom in the reactants is accounted for in the products. Balancing ensures the conservation of mass and atoms, a fundamental law of chemistry. In the context of neutralization reactions, writing and balancing the equation allows chemists to understand the stoichiometry involved. For example, in the neutralization reaction between H₂SO₄ and NaOH: - The balanced reaction is \[2 \mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2 \mathrm{H_2O} \]This tells us that two moles of NaOH are required to completely react with one mole of H₂SO₄. The balanced equation shows how each component interacts, providing the stoichiometric coefficients necessary for calculating reactants and products during reactions. Understanding balanced equations is fundamental for performing accurate laboratory work, especially in titrations and reaction predictions.