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 \(0.500 \mathrm{M} \mathrm{NaOH}\) 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 HBr solution?

Short answer

The mass of \(Zn(OH)_2\) added to the HBr solution is calculated in 5 steps: 1. Calculate moles of HBr consumed in neutralization with NaOH: \(0.500\,mol/L \times \frac{88.5\,mL}{1000\,mL}\) 2. Calculate the initial moles of HBr: \(0.500\,mol/L \times 0.350\,L\) 3. Calculate moles of HBr that reacted with \(Zn(OH)_2\): \(initial\,moles\,of\,HBr - moles\,of\,HBr\,consumed\,in\,neutralization\,with\,NaOH\) 4. Calculate moles of \(Zn(OH)_2\): \(\frac{moles\,of\,HBr\,reacted\,with\,Zn(OH)_2}{2}\) 5. Calculate the mass of \(Zn(OH)_2\): \(moles\,of\,Zn(OH)_2 \times molar\,mass\,of\,Zn(OH)_2\) Following these steps, the mass of \(Zn(OH)_2\) added to the HBr solution can be found.

Step-by-step solution

Calculate moles of HBr consumed in neutralization with NaOH

We know that the titration was performed using \(0.500\,M\) \(NaOH\) and it took \(88.5\,mL\) of the \(NaOH\) solution to fully neutralize the \(HBr\) in the solution. This means that the moles of \(HBr\) that were still left in the solution after reacting with the \(Zn(OH)_{2}\) can be calculated as follows: \[ moles\,of\,HBr = molarity\,of\,NaOH \times volume\,of\,NaOH = 0.500\,mol/L \times \frac{88.5\,mL}{1000\,mL} \]

Calculate the initial moles of HBr

We were given that \(0.350\,L\) of \(0.500\,M\) aqueous \(HBr\) was used initially. We can calculate the initial moles of \(HBr\) as follows: \[ initial\,moles\,of\,HBr = molarity\,of\,HBr \times volume\,of\,HBr = 0.500\,mol/L \times 0.350\,L \]

Calculate moles of HBr that reacted with Zn(OH)2

The difference between the initial moles of \(HBr\) and the moles of \(HBr\) consumed in the titration (neutralization with \(NaOH\)) will give us the moles of \(HBr\) that reacted with the \(Zn(OH)_2\). So, \[ moles\,of\,HBr\,reacted\,with\,Zn(OH)_2 = initial\,moles\,of\,HBr - moles\,of\,HBr\,consumed\,in\,neutralization\,with\,NaOH \]

Calculate moles of Zn(OH)2

The balanced equation for the reaction between \(Zn(OH)_2\) and \(HBr\) is: \[ Zn(OH)_2 + 2HBr \rightarrow ZnBr_2 + 2H_2O \] From the equation, we can see that 1 mole of \(Zn(OH)_2\) reacts with 2 moles of \(HBr\). Therefore, the moles of \(Zn(OH)_2\) can be calculated as follows: \[ moles\,of\,Zn(OH)_2 = \frac{moles\,of\,HBr\,reacted\,with\,Zn(OH)_2}{2} \]

Calculate the mass of Zn(OH)2

Now that we have calculated the moles of \(Zn(OH)_2\), we can calculate the mass of the \(Zn(OH)_2\) using its molar mass. The molar mass of \(Zn(OH)_2\) is approximately \(99.4\,g/mol\). The mass of the \(Zn(OH)_2\) can be calculated as follows: \[ mass\,of\,Zn(OH)_2 = moles\,of\,Zn(OH)_2 \times molar\,mass\,of\,Zn(OH)_2 \] Once we have the mass of \(Zn(OH)_2\), we have found the answer to the problem.

Key concepts

Understanding Neutralization Reactions

Neutralization reactions are fundamental to the study of chemistry as they occur when an acid and a base react to form a salt and water. This type of reaction is quintessential in titrations, where a known concentration of one reactant is used to find the concentration of an unknown. When dealing with titrations, it's crucial to comprehend the concept of the equivalence point, which is the exact point where the number of moles of acid equals the number of moles of base. In the given exercise, the neutralization reaction concerns the reaction between hydrobromic acid (HBr) and zinc hydroxide (Zn(OH)₂).

In a balanced neutralization reaction, the stoichiometric coefficients indicate the proportions in which the reactants combine. For instance, the reaction between HBr and Zn(OH)₂ can be represented as:
\[ Zn(OH)_2 + 2HBr \rightarrow ZnBr_2 + 2H_2O \]
Here, one mole of zinc hydroxide reacts with two moles of hydrobromic acid to produce one mole of zinc bromide and two moles of water. This stoichiometry is fundamental to solving the problem as it allows us to relate the moles of one reactant to the moles of another, leading us to determine unknown quantities in a chemical reaction.

Molar Mass Calculations

Calculating molar mass is a key step in carrying out stoichiometric calculations in chemical reactions. The molar mass is the mass of one mole of a given substance and is typically expressed in grams per mole (g/mol). It is equivalent to the atomic or molecular weight of the substance, based on the average weight of all isotopes present, and is numerically equal to the substance's relative formula mass. When given a compound, like zinc hydroxide in the exercise, you first need to determine the molar mass of each element within the compound and then sum them up according to the number of atoms of each element presented in the formula.

The molar mass of Zn(OH)₂ is calculated by adding the molar mass of zinc, which is approximately 65.4 g/mol, to twice the molar mass of oxygen, approximately 16 g/mol each, and to twice the molar mass of hydrogen, approximately 1 g/mol each. Then, you can deduce the molar mass to be:
\[Molar\,mass\,of\,Zn(OH)_2 = (1 \times 65.4) + (2 \times 16) + (2 \times 1) = 99.4\,g/mol\]
This molar mass allows us to convert between grams and moles, which is ultimately needed to determine the mass of the zinc hydroxide used in the reaction.

Stoichiometry of Chemical Reactions

Stoichiometry involves the quantitative relationship between the reactants and products in a chemical reaction. Understanding stoichiometry is pivotal for solving problems that ask for the mass of a substance when given volumes and concentrations. This involves a few key steps: (1) writing down the balanced chemical equation, (2) calculating the moles of each reactant and product based on the reaction's stoichiometry, and (3) using the moles with known quantities to find the unknowns. In this exercise, by employing stoichiometry, students are asked to derive the mass of Zn(OH)₂ based on the reaction with HBr.

The solution entails several conversions and uses the stoichiometry of the balanced equation. First, we determined the moles of HBr remaining after reacting with Zn(OH)₂ by titration with NaOH. Then, knowing the initial moles of HBr, we could subtract the moles that reacted with NaOH to get the moles that reacted with Zn(OH)₂. Given the reaction stoichiometry, for every one mole of Zn(OH)₂, two moles of HBr are required, leading us to establish the moles of zinc hydroxide. Multiplying by the molar mass of Zn(OH)₂ provides the final mass of the reactant. This logical sequence in calculating is essential in mastering stoichiometry in chemical reactions.