Question

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 $\mathrm{NaOH}$ 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

The mass of $\mathrm{Fe}(\mathrm{OH}{)}_3$ added to the ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution is $8.46\mathrm{g}$.

Step-by-step solution

Calculate the moles of H2SO4 in the initial solution

To calculate the moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in the $0.500\mathrm{L}$ of $0.250\mathrm{M}$ solution, we can use the formula: Moles = Molarity × Volume So, Moles of ${\mathrm{H}}_2{\mathrm{SO}}_4=0.250\mathrm{M}\times 0.500\mathrm{L}=0.125\mathrm{mol}$

Calculate the moles of unreacted H2SO4 remaining before titration with NaOH

To find the moles of unreacted ${\mathrm{H}}_2{\mathrm{SO}}_4$, we need to find the moles of ${\mathrm{OH}}^{-}$ ions present in the $12.5\mathrm{mL}$ of $0.500\mathrm{M}$ $\mathrm{NaOH}$. Moles of ${\mathrm{OH}}^{-}=0.500\mathrm{M}\times (12.5\times {10}^{-3})\mathrm{L}=0.00625\mathrm{mol}$ Since the ratio of ${\mathrm{H}}^{+}$ ions to ${\mathrm{OH}}^{-}$ ions in the neutralization reaction is 1:1, there must be the same number of moles of ${\mathrm{H}}^{+}$ ions provided by the unreacted ${\mathrm{H}}_{2}S{O}_4$. Moles of unreacted ${\mathrm{H}}_2{\mathrm{SO}}_4=0.00625\mathrm{mol}$

Calculate the moles of Fe(OH)3 that reacted with the initial H2SO4

In the reaction between $\mathrm{Fe}(\mathrm{OH}{)}_3$ and ${\mathrm{H}}_2{\mathrm{SO}}_4$, the ratio of $\mathrm{Fe}(\mathrm{OH}{)}_3$ to ${\mathrm{H}}_2{\mathrm{SO}}_4$ is 2:3. We initially had $0.125\mathrm{mol}$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$, and there are $0.00625\mathrm{mol}$ left unreacted. So, the reacted moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ are: Reacted moles of ${\mathrm{H}}_2{\mathrm{SO}}_4=0.125-0.00625=0.11875\mathrm{mol}$ Now, we can calculate the moles of $\mathrm{Fe}(\mathrm{OH}{)}_3$ that reacted with the initial ${\mathrm{H}}_2{\mathrm{SO}}_4$ using the 2:3 ratio: Moles of $\mathrm{Fe}(\mathrm{OH}{)}_3=(\frac{2}{3})\times 0.11875\mathrm{mol}=0.07917\mathrm{mol}$

Calculate the mass of Fe(OH)3

To calculate the mass of $\mathrm{Fe}(\mathrm{OH}{)}_3$, we use the formula: Mass = Moles × Molar mass The molar mass of $\mathrm{Fe}(\mathrm{OH}{)}_3$ is $106.87\mathrm{g}/\mathrm{mol}$. So, Mass of $\mathrm{Fe}(\mathrm{OH}{)}_3=0.07917\mathrm{mol}\times 106.87\mathrm{g}/\mathrm{mol}=8.46\mathrm{g}$ Therefore, the mass of $\mathrm{Fe}(\mathrm{OH}{)}_3$ added to the ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution is $8.46\mathrm{g}$.

Key concepts

Neutralization Reaction

Neutralization reactions occur when an acid reacts with a base to produce water and a salt. This type of reaction is a subset of chemical reactions, specifically involving acids and bases. The fundamental principle behind neutralization reactions is the interaction between hydrogen ions

• (${\mathrm{H}}^{+}$)
• hydroxide ions (${\mathrm{OH}}^{-}$)

The result is the formation of water (${\mathrm{H}}_2\mathrm{O}$). In the exercise provided, the solid $\mathrm{Fe}(\mathrm{OH}{)}_3$ represents the base, and ${\mathrm{H}}_2{\mathrm{SO}}_4$ is the acid. This particular reaction is slightly more complex because $\mathrm{Fe}(\mathrm{OH}{)}_3$ is a weak base and only partially dissociates, which is why not all of the ${\mathrm{H}}_2{\mathrm{SO}}_4$ was neutralized without adding additional base in the titration process.
The goal in a neutralization reaction is to balance the number of ${\mathrm{H}}^{+}$ and ${\mathrm{OH}}^{-}$ ions. If the amounts are perfectly balanced, the resulting solution will be neutral, i.e., neither acidic nor basic. In the case of this exercise, though ${\mathrm{H}}_2{\mathrm{SO}}_4$ was added in excess, excess ${\mathrm{H}}^{+}$ ions remained in the solution.

Titration

Titration is a method used in chemistry to determine the concentration of a solute in a solution. It involves the slow addition of a titrant from a burette into a measured volume of sample solution. The key to a successful titration is knowing the concentration of the titrant (standard solution) accurately. In this task, $\mathrm{NaOH}$ solution is used as the titrant to find out the amount of unreacted ${\mathrm{H}}_2{\mathrm{SO}}_4$ left in the solution.
Different types of titrations include:

• Acid-base titration: Targets the proton or hydroxide ion concentrations. This is the type used in our exercise, where $\mathrm{NaOH}$ (a base) reacts with ${\mathrm{H}}_2{\mathrm{SO}}_4$ (an acid).
• Redox titration: Involves electron transfer reactions.
• Complexometric titration: Applies to formation of a complex between metal ions.

In the scenario given, a specific volume (12.5 mL) of $\mathrm{NaOH}$ with a known molarity is used until the solution reaches an equivalence point. Titration helps you determine how much of the initial reagent (in our case, ${\mathrm{H}}_2{\mathrm{SO}}_4$) remains unreacted.

Molarity

Molarity, a central theme in chemistry, expresses the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. In formula terms, it is represented as:$$\text{Molarity}(M)=\frac{\text{Moles of solute}}{\text{Liters of solution}}$$This concept allows you to relate the amount of substance to its volume in solution and is vital for preparing solutions and performing quantitative chemical analysis, such as titration. In the given exercise, the molarity of the ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution and $\mathrm{NaOH}$ solution determines their role and quantity required in the reactions.

• ${\mathrm{H}}_2{\mathrm{SO}}_4$ has a molarity of 0.250 M.
• $\mathrm{NaOH}$, used in titration, is 0.500 M.

When you know the molarity, you can easily calculate the number of moles by simply multiplying the molarity by the volume of the solution. Knowledge of molarity enables chemists to predict how reactions proceed and understand their potential outputs.

Equivalence Point

The equivalence point in titration is the moment when the quantity of titrant added is sufficient to completely neutralize or react with the substance in solution. It signifies equal amounts of reactive species, such as when hydroxide ions and hydrogen ions neutralize each other exactly in stoichiometric proportions. However, the solution is not always neutral at this point, especially if strong acids or bases are involved and result in a salt that influences the solution’s pH.Reaching the equivalence point is critical in titration processes and was crucial in the exercise discussed. At the equivalence point,

• Moles of ${\mathrm{OH}}^{-}$ from $\mathrm{NaOH}$ equal the unreacted moles of ${\mathrm{H}}^{+}$ from the leftover ${\mathrm{H}}_2{\mathrm{SO}}_4$.

To detect the equivalence point, an indicator showing a visible change at the appropriate pH is often utilized, or potentiometry can be employed for more accuracy. Knowing when the equivalence point is reached allows us to compute back to the moles of reactants, facilitating calculations such as finding out how much $\mathrm{Fe}(\mathrm{OH}{)}_3$ was initially added in the solution in the exercise.