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Chapter 4: Problem 104

A solid sample of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is added to \(0.500 \mathrm{~L}\) of \(0.250 \mathrm{M}\) aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(12.5 \mathrm{~mL}\) of the NaOH solution to reach the equivalence point. What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) was added to the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Short Answer

Expert verified
The mass of \(\mathrm{Fe(OH)}_3\) added was approximately 8.68 g.

Step by step solution

01

Determine Initial Moles of Acid

First, calculate the moles of \(\mathrm{H}_2\mathrm{SO}_4\) in the initial solution:\[moles = Molarity \times Volume = 0.250\, \mathrm{M} \times 0.500\, \mathrm{L} = 0.125\, \mathrm{mol}\]
02

Find Moles of Base Used in Titration

Next, calculate the moles of \(\mathrm{NaOH}\) used in the titration:\[moles = Molarity \times Volume = 0.500\, \mathrm{M} \times 0.0125\, \mathrm{L} = 0.00625\, \mathrm{mol}\]This is the amount of moles that neutralized the excess acid.
03

Calculate Moles of Excess Acid

The moles of acidic protons neutralized by \(\mathrm{NaOH}\) is equivalent to the moles of \(\mathrm{NaOH}\) since 1 mole of \(\mathrm{NaOH}\) neutralizes 1 mole of \(\mathrm{H}^+\).Therefore, the excess moles of \(\mathrm{H}_2\mathrm{SO}_4\) after reacting with \(\mathrm{Fe(OH)}_3\) is:\[\mathrm{moles\ of\ } \mathrm{H}^+ = 0.00625\, \mathrm{mol}\]
04

Calculate Moles of Acid Reacted with \(\mathrm{Fe(OH)}_3\)

Since \(\mathrm{H}_2\mathrm{SO}_4\) is a diprotic acid, the total moles of protons is twice the moles of \(\mathrm{H}_2\mathrm{SO}_4\):\[0.125\, \mathrm{mol\ of\ } \mathrm{H}_2\mathrm{SO}_4 \times 2 = 0.250\, \mathrm{mol\ of\ protons}\]The moles of \(\mathrm{H}^+\) that reacted with \(\mathrm{Fe(OH)}_3\) is:\[0.250 - 0.00625 = 0.24375\, \mathrm{mol\ of\ } \mathrm{H}^+\]
05

Calculate Moles of \(\mathrm{Fe(OH)}_3\) Reacted

The reaction of \(\mathrm{Fe(OH)}_3\) with \(\mathrm{H}^+\) is:\[\mathrm{Fe(OH)}_3 + 3\, \mathrm{H}^+ \rightarrow \mathrm{Fe}^{3+} + 3\, \mathrm{H}_2\mathrm{O} \]Since 1 mole of \(\mathrm{Fe(OH)}_3\) reacts with 3 moles of \(\mathrm{H}^+\), the moles of \(\mathrm{Fe(OH)}_3\) is:\[\text{Moles of } \mathrm{Fe(OH)}_3 = \frac{0.24375}{3} = 0.08125\, \mathrm{mol}\]
06

Calculate Mass of \(\mathrm{Fe(OH)}_3\)

Finally, calculate the mass of \(\mathrm{Fe(OH)}_3\) using its molar mass (\(106.87\, \mathrm{g/mol}\)):\[\text{Mass} = 0.08125\, \mathrm{mol} \times 106.87\, \mathrm{g/mol} = 8.6786\, \mathrm{g}\]

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Key Concepts

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

Stoichiometry
Stoichiometry is like a recipe for chemists. It tells you how much of each ingredient, or reactant, you need to create a certain amount of product. This is based on the balanced chemical equation. In the given problem, we use stoichiometry to determine how much \( \mathrm{H}_2\mathrm{SO}_4 \) and \( \mathrm{Fe(OH)}_3 \) react with each other. The balanced chemical equation provides us the ratio of reactants and products.
This allows us to calculate the amounts involved in the reaction.
  • Through balanced equations, we find the mole-to-mole relationships.
  • Using these ratios, we convert between moles of different substances.
  • For example, 1 mole of \( \mathrm{Fe(OH)}_3 \) needs 3 moles of \( \mathrm{H}^+ \) ions for the reaction.
By understanding stoichiometry, we can also determine how much of the compound is needed or formed in a chemical reaction.
Molarity
Molarity is a measure of how concentrated a solution is. It is defined as the number of moles of solute per liter of solution, expressed in moles per liter (M). In this exercise, molarity helps us figure out how much \( \mathrm{H}_2\mathrm{SO}_4 \) was originally in the solution, and how much \( \mathrm{NaOH} \) was used in the titration.
  • The formula for molarity is: \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Liters of solution}} \]
  • Using this, we calculated that the \( \mathrm{H}_2\mathrm{SO}_4 \) solution had 0.125 moles.
  • It also confirmed that 0.00625 moles of \( \mathrm{NaOH} \) was used.
This step is essential to know how much reactant is available, which is crucial for stoichiometry and finding the mass of the compounds.
Chemical Reactions
Chemical reactions involve the transformation of substances. In our problem, \( \mathrm{Fe(OH)}_3 \) and\( \mathrm{H}_2\mathrm{SO}_4 \) undergo a reaction giving us water and iron ions in the process. During this titration process, another reaction occurs between \( \mathrm{H}_2\mathrm{SO}_4 \) and \( \mathrm{NaOH} \), which is crucial for determining the excess acid.
  • The original reaction: \( \mathrm{Fe(OH)}_3 + 3 \mathrm{H}^+ \rightarrow \mathrm{Fe}^{3+} + 3 \mathrm{H}_2\mathrm{O} \).
  • Shows that 1 mole of \( \mathrm{Fe(OH)}_3 \) reacts with 3 moles of \( \mathrm{H}^+ \).
  • Through titration, we determine how much \( \mathrm{H}^+ \) was neutralized by \( \mathrm{NaOH} \).
Understanding these reactions allows us to calculate the amounts of substances that reacted or remained unreacted. Thus, it helps us complete the stoichiometry calculation successfully.
Diprotic Acids
Diprotic acids are acids that can donate two protons or hydrogen ions per molecule to a solution. Sulfuric acid \( \mathrm{H}_2\mathrm{SO}_4 \) is an example of a diprotic acid. This means each mole of sulfuric acid can potentially release two moles of \( \mathrm{H}^+ \) ions. Knowing that \( \mathrm{H}_2\mathrm{SO}_4 \) is diprotic is crucial for this problem.
  • Initial calculation for \( \mathrm{H}_2\mathrm{SO}_4 \): \[ 0.125 \, \text{mol} \, \mathrm{H}_2\mathrm{SO}_4 \times 2 = 0.250 \, \text{mol} \, \mathrm{H}^+ \]
  • This informs us of the total acidic potential of the solution.
  • This concept feeds into the stoichiometric calculations that follow.
By grasping the concept of diprotic acids, we can accurately consider their involvement and contribution to the overall chemical reaction.

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Most popular questions from this chapter

A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is \(1.5 \times 10^{-9} \mathrm{M},\) and \(1.00 \mathrm{~mL}\) of this solution will be delivered to a dish containing \(2.0 \times 10^{5}\) cancer cells in \(5.00 \mathrm{~mL}\) of aqueous fluid. What is the ratio of drug molecules to the number of cancer cells in the dish?

State whether each of the following statements is true or false. Justify your answer in each case. (a) \(\mathrm{NH}_{3}\) contains no OH \(^{-}\) ions, and yet its aqueous solutions are basic. (b) HF is a strong acid. (c) Although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reac- tion. (a) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{NaOH},(\mathbf{b}) \mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{FeCl}_{2}\)

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out \(\mathrm{AgCl}(s) .\) What volume of a \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (b) You could add solid \(\mathrm{KCl}\) to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution costs \(\$ 39.95\) for \(500 \mathrm{~mL}\) and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

Suppose you have \(3.00 \mathrm{~g}\) of powdered zinc metal, \(3.00 \mathrm{~g}\) of powdered silver metal and \(500.0 \mathrm{~mL}\) of a \(0.2 \mathrm{M}\) copper(II) nitrate solution. (a) Which metal will react with the copper(II) nitrate solution? (b) What is the net ionic equation that describes this reaction? (c) Which is the limiting reagent in the reaction? (d) What is the molarity of \(\mathrm{Cu}^{2+}\) ions in the resulting solution?
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