Problem 104 - Chemistry The Central Science

Question

A solid sample of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to 0.350 \(\mathrm{L}\) of 0.500 \(\mathrm{M}\) aqueous HBr. The solution that remains is still acidic. It is then titrated with 0.500 \(\mathrm{MNaOH}\) solution, and it takes 88.5 mL of the NaOH solution to reach the equivalence point. What mass of \(\mathrm{Zn}(\mathrm{OH})_{2}\) was added to the HBr solution?

Step-by-Step Solution

Answer

The mass of Zn(OH)_2 added to the HBr solution is approximately \(2.20 \thinspace g\).

Step 1: Calculate the number of moles of HBr initially present

To find the number of moles of HBr initially present in the solution, multiply the volume of the HBr solution in liters by the molarity of the solution: n(HBr_initial) = volume (L) × molarity n(HBr_initial) = (0.350 L) × (0.500 mol/L) = 0.175 mol

Step 2: Balanced chemical equation for the reaction between Zn(OH)_2 and HBr

The balanced chemical equation for the reaction between Zn(OH)_2 and HBr is: Zn(OH)_2 + 2HBr → ZnBr_2 + 2H_2O

Step 3: Calculate the number of moles of HBr reacted with Zn(OH)_2

Since it takes 88.5 mL of 0.500 M NaOH solution to reach the equivalence point, we can calculate the number of moles of NaOH and hence, the number of moles of HBr reacted with Zn(OH)_2. n(NaOH) = volume(L) × molarity n(NaOH) = (88.5 mL)(1 L/1000 mL) × (0.500 mol/L) = 0.04425 mol From the balanced chemical equation, it takes 2 moles of HBr to react with 1 mole of Zn(OH)_2. Therefore, the number of moles of HBr that reacted with Zn(OH)_2 is the same as the number of moles of NaOH used in the titration: n(HBr_reacted) = n(NaOH) = 0.04425 mol

Step 4: Calculate the number of moles of Zn(OH)_2 added

From the balanced chemical equation, we know that 2 moles of HBr react with 1 mole of Zn(OH)_2. Thus, we can find the number of moles of Zn(OH)_2 added by dividing the number of moles of HBr that reacted with Zn(OH)_2 by 2: n(Zn(OH)_2) = n(HBr_reacted) / 2 = 0.04425 mol / 2 = 0.022125 mol

Step 5: Determine the mass of Zn(OH)_2 added

Now that we have the number of moles of Zn(OH)_2 added, we can calculate the mass of Zn(OH)_2 by using its molar mass (Zn = 65.38 g/mol, O = 16.00 g/mol, H = 1.01 g/mol): Molar mass of Zn(OH)_2 = 65.38 + 2(16.00 + 1.01) = 65.38 + 2(17.01) = 99.40 g/mol So, the mass of Zn(OH)_2 added is: mass(Zn(OH)_2) = n(Zn(OH)_2) × molar mass(Zn(OH)_2) mass(Zn(OH)_2) = 0.022125 mol × 99.40 g/mol = 2.19825 g Thus, the mass of Zn(OH)_2 added to the HBr solution is approximately \(2.20 \thinspace g\).

Key Concepts

Molarity Calculation

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This concept is crucial in chemistry because it allows for the quantitative analysis of substances in a mixture. In our exercise, molarity was used to determine the amount of acid (HBr) and base (NaOH) present in the given volumes of their solutions. To calculate molarity, we use the formula:

\[\begin{equation} M = \frac{n}{V} \end{equation}\]
where:\

\
• \(M\) is the molarity of the solution,\
• \(n\) is the number of moles of the solute,\
• \(V\) is the volume of the solution in liters.\

\
For instance, if we have 0.175 moles of HBr in a 0.350 L solution, the molarity would be calculated as \(0.500 \frac{mol}{L}\), confirming the initial concentration given in our problem.

Stoichiometry

Stoichiometry is the section of chemistry that involves using relationships from balanced chemical equations to calculate quantities of reactants and products. It's like a recipe: if you know how much of one ingredient you have, you can figure out how much of the others you need or will produce. In our exercise, we used stoichiometry to relate the moles of HBr that reacted with Zn(OH)_2 to the moles of NaOH needed to reach the equivalence point in a titration. Because the ratio of HBr to Zn(OH)_2 is 2:1, we can deduce that half the moles of HBr reacted will equal the moles of Zn(OH)_2 added. Such stoichiometric relationships spring from a balanced chemical equation, which serves as the foundation for these calculations.

Balanced Chemical Equation

A balanced chemical equation provides the proportions of reactants and products in a chemical reaction. It ensures the law of conservation of mass is obeyed - matter can neither be created nor destroyed. The coefficients in a balanced equation tell us the ratio in which substances react and are formed. For our exercise, the relevant equation was:

\[\begin{equation}Zn(OH)_2 + 2HBr \rightarrow ZnBr_2 + 2H_2O\end{equation}\]
This equation indicates that one mole of Zn(OH)_2 reacts with two moles of HBr to form one mole of ZnBr_2 and two moles of water. Using this balanced equation, we were able to deduce the molar relationship between our reactants and products, pivotal for the stoichiometric calculations in our titration problem.

Mole Concept

The mole concept is a bridge between the macroscopic world we observe and the microscopic world of atoms and molecules. One mole is 6.022 x 10^23 (Avogadro's number) of particles, whether they're atoms, ions, or molecules. In our practice problem, we had to first calculate the number of moles of Zn(OH)_2. Understanding that one mole of a substance contains Avogadro's number of that substance enables us to convert between the mass of a substance and the number of moles using the formula:

\[\begin{equation}n = \frac{m}{M}\end{equation}\]
where:\

\
• \(n\) is the number of moles,\
• \(m\) is the mass of the substance,\
• \(M\) is the molar mass of the substance.\

\
By applying this concept, we converted the calculated moles of Zn(OH)_2 to grams to find the mass of Zn(OH)_2 added to the solution, which is an essential skill for any chemistry student tackling reactions and stoichiometry.