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Mobile App Chapter 17: Problem 74
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
Expert verified
The mass of BaSO4 formed is 9.97 grams and the pH of the mixed solution is 13.03.
Step by step solution
01
Identify the balanced chemical equation
The reaction between Ba(OH)2 and H2SO4 forms BaSO4 and water. Hence, the balanced chemical equation is: \[Ba(OH)2 + H2SO4 \rightarrow BaSO4 + 2H2O\]. This means, one mole of Ba(OH)2 reacts with one mole of H2SO4 to form one mole of BaSO4.
02
Calculate moles of each reactant
We can calculate the moles of reactants using the formula:\[ Moles = Molarity \times Volume(L)\].For Ba(OH)2, moles = \(1.00 M \times 0.050 L = 0.05 moles\)For H2SO4, moles = \(0.494 M \times 0.0864 L = 0.0427 moles\)
03
Identify the limiting reagent
H2SO4 is the limiting reagent because it has fewer moles (0.0427 moles) than Ba(OH)2 (0.05 moles). Thus, H2SO4 will completely react and determine the maximum amount of BaSO4 formed.
04
Calculate the mass of BaSO4 formed
The molar mass of BaSO4 is approximately 233.43 g/mole. The amount of BaSO4 formed is determined by the moles of the limiting reactant as one mole of H2SO4 produces one mole of BaSO4. Therefore, the mass of BaSO4 formed = moles of limiting reagent \times molar mass of BaSO4 = \(0.0427 moles \times 233.43 g/mole = 9.97 grams\) of BaSO4.
05
Calculate the remaining moles of Ba(OH)2
The remaining moles of Ba(OH)2 are calculated by subtracting the moles of H2SO4 (limiting reagent) from the initial moles of Ba(OH)2. Hence, remaining moles of Ba(OH)2 = \(0.050 moles - 0.0427 moles = 0.0073 moles\). This remaining Ba(OH)2 affects the pH of the solution.
06
Calculate the concentration of Ba(OH)2 in the solution
The final volume of the solution is the sum of the volumes of Ba(OH)2 and H2SO4, which is \(0.050 L + 0.0864 L = 0.1364 L\). The concentration of Ba(OH)2 in the solution is given by \( \frac{moles of Ba(OH)2}{final volume in L} = \frac{0.0073 moles}{0.1364 L} = 0.0535 M\).
07
Calculate the pH of the solution
Knowing that Ba(OH)2 is a strong base, it will ionize completely in solution. For every mole of Ba(OH)2, two moles of OH- ions are formed. Thus concentration of OH- in the final solution = 2 \times concentration of Ba(OH)2 = \(2 \times 0.0535 M = 0.107 M\). The pOH of the solution can be calculated using the formula \(-\log[OH^-]\), and then subtracted the pOH from 14 to find the pH. Hence, pOH = -log(0.107) = 0.97 and therefore, pH = 14 - pOH = 14 - 0.97 = 13.03.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limiting Reagent Calculation
The concept of a limiting reagent is essential in chemical reaction stoichiometry, as it influences how much product can be formed in a reaction. It refers to the substance that is entirely used up first during a chemical reaction, thus determining when the reaction stops. When two or more reactants are mixed, there will often be a reactant that is not in excess; this reactant is known as the limiting reagent.
To identify the limiting reagent, we start by calculating the moles of each reactant involved in the reaction. This involves the use of molarity (the number of moles of solute per liter of solution) and the volume of the solutions mixed. After the moles are determined, the stoichiometry of the balanced chemical equation is used as a guide. For example, in the provided exercise, the chemical equation shows a 1:1 ratio between Ba(OH)2 and H2SO4. By comparing the moles calculated to the stoichiometry, the limiting reagent is identified, which in this case is H2SO4. For the maximum yield of a product, the amount of product formed is always based on the moles of the limiting reagent.
To identify the limiting reagent, we start by calculating the moles of each reactant involved in the reaction. This involves the use of molarity (the number of moles of solute per liter of solution) and the volume of the solutions mixed. After the moles are determined, the stoichiometry of the balanced chemical equation is used as a guide. For example, in the provided exercise, the chemical equation shows a 1:1 ratio between Ba(OH)2 and H2SO4. By comparing the moles calculated to the stoichiometry, the limiting reagent is identified, which in this case is H2SO4. For the maximum yield of a product, the amount of product formed is always based on the moles of the limiting reagent.
Molarity and Volume Relationship
Molarity is the concentration of a solute in a solution, and it represents the number of moles of a substance per liter of solution. It is expressed in moles per liter (M). The relationship between molarity (M), volume (V), and moles (n) of a solute is given by the equation:
\[ n = M \times V \]
where n is the number of moles, M is the molarity, and V is the volume in liters. In the context of the given exercise, we utilized this relationship to find the moles of Ba(OH)2 and H2SO4 by multiplying the molarity by the volume. Understanding this relationship is crucial because it connects the volumes and concentrations of reactants in a reaction to their respective moles, which are used to determine the stoichiometry and the limiting reagent.
\[ n = M \times V \]
where n is the number of moles, M is the molarity, and V is the volume in liters. In the context of the given exercise, we utilized this relationship to find the moles of Ba(OH)2 and H2SO4 by multiplying the molarity by the volume. Understanding this relationship is crucial because it connects the volumes and concentrations of reactants in a reaction to their respective moles, which are used to determine the stoichiometry and the limiting reagent.
pH Calculation of Strong Base
The pH scale is a measure of the acidity or basicity of an aqueous solution. The pH of a solution is a logarithmic representation of the concentration of hydrogen ions (\( H^+ \)), with lower values being more acidic and higher values being more basic. For a strong base, such as Ba(OH)2, it completely ionizes in solution to yield hydroxide ions (\( OH^- \)).
To find the pH, one must first calculate the concentration of the hydroxide ions produced by the strong base. This step involves doubling the molarity of the strong base, as each mole of Ba(OH)2 produces two moles of OH-. Once we have the concentration of OH-, the pOH is calculated using the formula:
\[ \text{pOH} = -\log[OH^-] \]
The pH can then be found by subtracting the pOH from 14 (since \( \text{pH} + \text{pOH} = 14 \) for any aqueous solution at 25°C). This process was applied in the exercise to determine the pH of the mixed solution, resulting in a basic pH of 13.03, characteristic of an environment with excess hydroxide ions.
To find the pH, one must first calculate the concentration of the hydroxide ions produced by the strong base. This step involves doubling the molarity of the strong base, as each mole of Ba(OH)2 produces two moles of OH-. Once we have the concentration of OH-, the pOH is calculated using the formula:
\[ \text{pOH} = -\log[OH^-] \]
The pH can then be found by subtracting the pOH from 14 (since \( \text{pH} + \text{pOH} = 14 \) for any aqueous solution at 25°C). This process was applied in the exercise to determine the pH of the mixed solution, resulting in a basic pH of 13.03, characteristic of an environment with excess hydroxide ions.
