Problem 104
Question
A solid sample of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(0.350 \mathrm{~L}\) of \(0.500 M\) aqueous HBr. The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(88.5 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Zn}(\mathrm{OH})_{2}\) was added to the \(\mathrm{HBr}\) solution?
Step-by-Step Solution
Verified Answer
The mass of Zn(OH)₂ added to the HBr solution is \(6.50\:g\).
1Step 1: Calculate initial moles of HBr
To determine the initial moles of HBr, use the initial volume and concentration given in the problem.
\[ moles\: of \:HBr = volume \times concentration \]
\[ moles\: of \:HBr = 0.350\: L \times 0.500\: M \]
\[ moles\: of \:HBr = 0.175\: moles \]
2Step 2: Determine moles of HBr that reacted with NaOH during titration
To determine the moles of HBr that reacted with the NaOH solution, use the volume and concentration of NaOH given in the problem.
\[ moles\: of \:HBr\: reacted\: with\: NaOH = volume \times concentration \]
\[ moles\: of \:HBr\: reacted\: with\: NaOH = 0.0885\: L \times 0.500\: M \]
\[ moles\: of \:HBr\: reacted\: with\: NaOH = 0.04425\: moles \]
3Step 3: Calculate moles of HBr that reacted with Zn(OH)₂
Subtract the moles of HBr that reacted with the NaOH solution from the initial moles of HBr to determine how many moles of HBr reacted with the solid Zn(OH)₂.
\[moles\: of \: HBr\: reacted\: with\: Zn(OH)₂ = initial\: moles\: of\: HBr - moles\: of \: HBr\: reacted\: with\: NaOH \]
\[moles\: of \: HBr\: reacted\: with\: Zn(OH)₂ = 0.175 - 0.04425 \]
\[moles\: of \: HBr\: reacted\: with\: Zn(OH)₂ = 0.13075 \: moles \]
4Step 4: Calculate moles of Zn(OH)₂
The balanced chemical equation for the reaction between Zn(OH)₂ and HBr is:
\[Zn(OH)₂(s) + 2HBr(aq) \rightarrow ZnBr₂(aq) + 2H₂O(l)\]
From the stoichiometry of the reaction, 1 mole of Zn(OH)₂ reacts with 2 moles of HBr. We can use this relationship to determine the moles of Zn(OH)₂.
\[ moles\: of \: Zn(OH)₂ = \frac{moles\: of \: HBr\: reacted\: with\: Zn(OH)₂}{2} \]
\[ moles\: of \: Zn(OH)₂ = \frac{0.13075\: moles}{2} \]
\[ moles\: of \: Zn(OH)₂ = 0.065375 \: moles \]
5Step 5: Calculate mass of Zn(OH)₂
Use the molar mass of Zn(OH)₂ (99.42 g/mol) to convert moles of Zn(OH)₂ to grams.
\[ mass\: of \: Zn(OH)₂ = moles \times molar\: mass \]
\[ mass\: of \: Zn(OH)₂ = 0.065375\: moles \times 99.42 g/mol \]
\[ mass\: of \: Zn(OH)₂ = 6.50\:g \]
Thus, the mass of Zn(OH)₂ added to the HBr solution is 6.50 grams.
Key Concepts
Titration CalculationsStoichiometryMolarityChemical Reaction Equation
Titration Calculations
Titration is a technique used in chemistry to determine the concentration of an unknown substance by reacting it with a standard solution of known concentration. This process involves the gradual addition of the standard solution into the unknown solution until the reaction reaches the equivalence point—the point at which the reactants are in stoichiometric proportion and the reaction is complete.
During the titration process, we often measure volumes using a burette and use indicators such as a pH meter or color-changing chemicals to detect the equivalence point. In our exercise, we analyzed how much 0.500 M NaOH was needed to neutralize the remaining HBr in a solution, indicating that we had reached the equivalence point. The volume of NaOH used in the titration provides key information for calculating the amount of material in the original solution.
During the titration process, we often measure volumes using a burette and use indicators such as a pH meter or color-changing chemicals to detect the equivalence point. In our exercise, we analyzed how much 0.500 M NaOH was needed to neutralize the remaining HBr in a solution, indicating that we had reached the equivalence point. The volume of NaOH used in the titration provides key information for calculating the amount of material in the original solution.
Stoichiometry
Stoichiometry is a section of chemistry that involves using balanced chemical equations to calculate the relationships between the quantities of reactants and products in a chemical reaction.
For our titration problem, we used the balanced chemical equation
\[\text{Zn(OH)}_{2}(s) + 2\text{HBr}(aq) \rightarrow \text{ZnBr}_{2}(aq) + 2\text{H}_{2}\text{O}(l)\] which told us that 2 moles of HBr react with 1 mole of \(\text{Zn(OH)}_{2}\). This stoichiometric relationship is the foundation for all subsequent calculations in the problem, as it allows us to connect the moles of HBr reacted with the moles of \(\text{Zn(OH)}_{2}\) that were initially added.
For our titration problem, we used the balanced chemical equation
\[\text{Zn(OH)}_{2}(s) + 2\text{HBr}(aq) \rightarrow \text{ZnBr}_{2}(aq) + 2\text{H}_{2}\text{O}(l)\] which told us that 2 moles of HBr react with 1 mole of \(\text{Zn(OH)}_{2}\). This stoichiometric relationship is the foundation for all subsequent calculations in the problem, as it allows us to connect the moles of HBr reacted with the moles of \(\text{Zn(OH)}_{2}\) that were initially added.
Molarity
Molarity, often represented as M, is a measure of concentration in chemistry that indicates the number of moles of solute per liter of solution. It's a crucial concept in titrations since it allows us to link the volume of a solution to the number of moles of a solute it contains.
The problem provided the molarity of both the HBr and NaOH solutions (0.500 M), which we used to determine the moles of each reactant. Understanding molarity is essential for titration calculations, as it enables chemists to accurately prepare solutions with precise concentrations needed for a variety of reactions.
The problem provided the molarity of both the HBr and NaOH solutions (0.500 M), which we used to determine the moles of each reactant. Understanding molarity is essential for titration calculations, as it enables chemists to accurately prepare solutions with precise concentrations needed for a variety of reactions.
Chemical Reaction Equation
A chemical reaction equation provides a symbolic representation of a chemical reaction, showing the reactants and products along with their quantities in moles. In our example, the balanced chemical equation clearly displays how zinc hydroxide reacts with hydrobromic acid to form zinc bromide and water.
Chemical equations must be balanced, meaning that the number of atoms of each element on the reactant side must equal the number of atoms of that element on the product side. This balancing act is known as the law of conservation of mass. The equation provided the necessary framework for performing stoichiometric calculations, which is integral to solving this titration problem.
Chemical equations must be balanced, meaning that the number of atoms of each element on the reactant side must equal the number of atoms of that element on the product side. This balancing act is known as the law of conservation of mass. The equation provided the necessary framework for performing stoichiometric calculations, which is integral to solving this titration problem.
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