Question

A student mixes \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(86.4 \mathrm{~mL}\) of \(0.494 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) Calculate the mass of \(\mathrm{BaSO}_{4}\) formed and the \(\mathrm{pH}\) of the mixed solution.

Short answer

9.97 g of \( \mathrm{BaSO}_{4} \) is formed; pH is 12.73.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction between \( \mathrm{Ba(OH)}_{2} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) can be written as:\[ \mathrm{Ba(OH)}_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4} + 2\mathrm{H}_2\mathrm{O} \]This equation shows that one mole of \( \mathrm{Ba(OH)}_{2} \) reacts with one mole of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) to produce one mole of \( \mathrm{BaSO}_{4} \) and two moles of water.

Calculate Moles of Reactants

To find the moles of each reactant, use the formula:\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]For \( \mathrm{Ba(OH)}_{2} \): \[ 1.00 \mathrm{M} \times 0.0500 \mathrm{~L} = 0.0500 \text{ moles} \]For \( \mathrm{H}_{2} \mathrm{SO}_{4} \): \[ 0.494 \mathrm{M} \times 0.0864 \mathrm{~L} = 0.0427 \text{ moles} \]

Determine the Limiting Reactant

Since the balanced equation shows a 1:1 ratio, compare the moles of each reactant. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) has fewer moles (0.0427) than \( \mathrm{Ba(OH)}_{2} \) (0.0500), so \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is the limiting reactant.

Calculate the Mass of \( \mathrm{BaSO}_{4} \) Formed

Using the limiting reactant's moles, calculate the moles of \( \mathrm{BaSO}_{4} \) formed (equal to moles of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) because of 1:1 stoichiometry):\[ 0.0427 \text{ moles of } \mathrm{BaSO}_{4} \]To find this mass, use the molar mass of \( \mathrm{BaSO}_{4} \) (233.39 g/mol):\[ 0.0427 \text{ moles} \times 233.39 \text{ g/mol} = 9.97 \text{ grams} \]

Calculate the Remaining \( \mathrm{Ba(OH)}_{2} \) Moles

Since \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is the limiting reactant, it completely reacts. Subtract the used moles from the initial moles of \( \mathrm{Ba(OH)}_{2} \):\[ 0.0500 - 0.0427 = 0.0073 \text{ moles of } \mathrm{Ba(OH)}_{2} \text{ remain} \]

Calculate \( \mathrm{pH} \) of the Solution

The remaining \( \mathrm{Ba(OH)}_{2} \) will contribute additional hydroxide ions, thus affecting the pH. Calculate the concentration of remaining OH⁻ ions:\[ [\mathrm{OH}^{-}] = \frac{0.0073 \text{ moles}}{0.1364 \text{ L}} = 0.0535 \text{ M} \]The \( \mathrm{pOH} \) is:\[ \mathrm{pOH} = -\log(0.0535) = 1.27 \]Given the relation \( \mathrm{pH} = 14 - \mathrm{pOH} \):\[ \mathrm{pH} = 14 - 1.27 = 12.73 \]

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, halting the reaction since no further products can form without it. Identifying the limiting reactant is crucial because it determines the maximum amount of product that a reaction can produce.
In the exercise at hand, we have a reaction between barium hydroxide, \( \mathrm{Ba(OH)}_{2} \), and sulfuric acid, \( \mathrm{H}_{2} \mathrm{SO}_{4} \). The provided amounts are 0.0500 moles of \( \mathrm{Ba(OH)}_{2} \) and 0.0427 moles of \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

• The balanced equation shows a 1:1 stoichiometric ratio between these reactants.
• To find the limiting reactant, compare the available moles against this ratio.
• Since \( \mathrm{H}_{2} \mathrm{SO}_{4} \) has fewer moles than \( \mathrm{Ba(OH)}_{2} \), it is the limiting reactant.

In summary, the limiting reactant controls how much \( \mathrm{BaSO}_{4} \) will form, and it is always the starting point in calculating the yield of products.

Balanced Chemical Equation

A balanced chemical equation accurately represents the conservation of mass in a reaction, where the number of atoms for each element is the same on both sides of the equation. Balancing ensures that no atoms are lost or gained, only rearranged.
For example, the equation for the reaction in this exercise is:\[ \mathrm{Ba(OH)}_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4} + 2 \mathrm{H}_2 \mathrm{O} \]Here, each side of the equation contains:

• 1 barium (Ba) atom
• 2 hydroxide (OH) ions
• 1 sulfur (S) atom
• 4 oxygen (O) atoms in sulfate, plus 2 in hydroxide and 2 in water, balancing to 10 oxygen atoms
• 4 hydrogen (H) atoms in both water and sulfuric acid

By ensuring that both sides match in terms of atoms, this balanced equation helps in calculating reactant and product amounts accurately, paving the way for solving limiting reactant and product formation issues.

pH Calculation

pH is a measure of the hydrogen ion concentration in a solution, reflecting its acidity or basicity. In this scenario, we must consider the excess of \( \mathrm{Ba(OH)}_{2} \) after the limiting reactant \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is consumed.
The remaining hydroxide ions from \( \mathrm{Ba(OH)}_{2} \) affect the solution's basicity. To find the pH:

• First, calculate the remaining moles of \( \mathrm{Ba(OH)}_{2} \): 0.0073 moles.
• Determine the concentration of \( \mathrm{OH}^{-} \) ions: 0.0535 M.
• The \( \mathrm{pOH} \) is then calculated as \( \mathrm{pOH} = -\log(0.0535) \).
• Finally, convert \( \mathrm{pOH} \) to \( \mathrm{pH} \) using the relation \( \mathrm{pH} = 14 - \mathrm{pOH} \).

In this case, the resulting \( \mathrm{pH} \) value reflects a basic solution due to the leftover hydroxide ions, derived from the excess \( \mathrm{Ba(OH)}_{2} \). Understanding these steps is crucial in applications involving reactions that produce or use hydroxide ions.

Molar Mass

Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It allows chemists to convert between the number of moles and the mass of a substance.
For \( \mathrm{BaSO}_{4} \), calculating the molar mass involves adding the atomic masses of its component atoms:

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol each, for a total of 64.00 g/mol (4 atoms)

Summing these gives a molar mass of 233.39 g/mol for \( \mathrm{BaSO}_{4} \).
This value allows us to translate moles of \( \mathrm{BaSO}_{4} \) into grams, using the equation:\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]For example, 0.0427 moles of \( \mathrm{BaSO}_{4} \) converts to approximately 9.97 grams, illustrating how knowing molar mass is fundamental in stoichiometry for determining product yields.