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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 \(\mathrm{it}\) 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

Expert verified
The mass of the Zn(OH)鈧 sample added to the HBr solution is approximately 6.50 g.

Step by step solution

01

Calculate the moles of \(\mathrm{H}^+\) ions left in the HBr solution

Using the volume and molarity of the \(\mathrm{NaOH}\) solution used for titration, we can calculate the moles of remaining \(\mathrm{H}^+\) ions in the HBr solution after reacting with \(\mathrm{Zn(OH)_2}\). Moles of NaOH = Molarity 脳 Volume Moles of NaOH = 0.500 M 脳 \(\frac{88.5 mL}{1000 mL/L}\) = 0.04425 moles Since the moles of NaOH and H鈦 react 1:1 to produce water (according to the neutralization reaction between a strong acid and a strong base), the moles of remaining H鈦 ions can be determined: Moles of H鈦 ions remaining in HBr solution = 0.04425 moles
02

Calculate the moles of \(\mathrm{Zn(OH)_2}\) that reacted with HBr

Now we need to find out how many moles of H鈦 ions were in the HBr solution initially. Moles of H鈦 ions in HBr solution initially = Molarity 脳 Volume Moles of H鈦 ions in HBr solution initially = 0.500 M 脳 0.350 L = 0.175 moles From this, we can determine the moles of H鈦 ions that reacted with Zn(OH)鈧: Moles of H鈦 ions reacted with Zn(OH)鈧 = Initial moles of H鈦 ions - Remaining moles of H鈦 ions Moles of H鈦 ions reacted with Zn(OH)鈧 = 0.175 moles - 0.04425 moles = 0.13075 moles Each molecule of Zn(OH)鈧 reacts with two H鈦 ions, so we can calculate the moles of Zn(OH)鈧 that reacted with HBr: Moles of Zn(OH)鈧 = Moles of H鈦 ions reacted with Zn(OH)鈧 梅 2 Moles of Zn(OH)鈧 = 0.13075 moles 梅 2 = 0.065375 moles
03

Calculate the mass of \(\mathrm{Zn(OH)_2}\)

Using the molar mass of Zn(OH)鈧 (65.38 g/mol for Zn + 2脳(15.999 g/mol for O + 1.008 g/mol for H) = 99.394 g/mol), we can calculate the mass of Zn(OH)鈧: Mass of Zn(OH)鈧 = Moles of Zn(OH)鈧 脳 Molar mass of Zn(OH)鈧 Mass of Zn(OH)鈧 = 0.065375 moles 脳 99.394 g/mol 鈮 6.50 g The mass of the Zn(OH)鈧 sample added to the HBr solution is approximately 6.50 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Titration
In the chemistry lab, titration is a crucial analytical method used to determine the concentration of a known reactant. Acid-base titration, in particular, involves the gradual addition of an acid to a base (or vice versa) to neutralize the pH of the solution.

The point at which the acid completely neutralizes the base (or vice versa) is known as the equivalence point. It鈥檚 usually marked by a sudden change in the color of the indicator present in the solution, indicating that stoichiometrically equivalent amounts of acid and base have reacted.

When working acid-base titration problems, it鈥檚 important to remember the neutralization reaction typically is a 1:1 reaction, where one mole of acid reacts with one mole of base to produce salt and water. However, it is essential to consider the exact reaction taking place, as some acids and bases can react in different molar ratios. Always write out the balanced equation to guide your stoichiometric calculations.
Stoichiometry
Stoichiometry stems from the Greek words 'stoicheion' (element) and 'metron' (measure), and it's a section of chemistry that involves calculating the amounts of reactants and products in a chemical reaction.

It is the foundation for many procedures in chemistry, such as determining concentrations in titrations. Stoichiometry requires a balanced chemical equation to ensure that mass and charge are conserved and provides the framework to understand the relationship between different substances in a reaction.

The stoichiometry of a reaction can tell us how many moles of a reactant are needed to react with a certain amount of another reactant. We can use this information to find unknown concentrations, predict the amount of product formed, or in the case of our exercise, determine the amount of an acid necessary to neutralize a base, or vice versa.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance (the Avogadro number of molecules), and it's expressed in grams per mole (g/mol). It's a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters that we interact with in the lab.

To calculate molar mass, you simply add up the atomic masses for each element in the molecule from the periodic table. For instance, zinc hydroxide, \( \mathrm{Zn(OH)_2} \), consists of one zinc atom, two oxygen atoms, and two hydrogen atoms.

The molar mass calculation is essential in stoichiometry calculations, as seen in the provided exercise. By knowing the molar mass, you can convert between mass and moles of a substance鈥攅nabling you to precisely measure out reactants for a reaction or, in the case of our zinc hydroxide sample, determine the mass of a substance given its quantity in moles.

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Most popular questions from this chapter

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the observations made above. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with 3.00 \(\mathrm{L}\) of \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

(a) How many grams of solute are present in \(50.0 \mathrm{~mL}\) of \(0.488 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (b) If \(4.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in enough water to form \(400 \mathrm{~mL}\) of solution, what is the molarity of the solution? (c) How many milliliters of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) contain \(1.75 \mathrm{~g}\) of solute?

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) \(3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M}\) HCl. The \(\mathrm{NH}_{3}\) reacts with HCl as follows: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min}\), the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) ls this manufacturer in compliance with regulations?

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