Question

14..How many milliliters of \(0.05 \mathrm{M} \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) solution are required for titration of \(60 \mathrm{ml}\) of \(0.01\) M \(\mathrm{ZnSO}_{4}\) solution, when the product of reaction is \(\mathrm{K}_{2} \mathrm{Zn}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2} ?\)

Short answer

8 mL of 0.05 M \(\text{K}_4[\text{Fe}(\text{CN})_6]\) solution are required.

Step-by-step solution

Write the Balanced Chemical Equation

First, we need to identify the balanced chemical equation for the reaction. The given reactant solutions are \( ext{K}_4[ ext{Fe}( ext{CN})_6]\) and \( ext{ZnSO}_4\). The product is \( ext{K}_2 ext{Zn}_3[ ext{Fe}( ext{CN})_6]_2\). The equation is:\[3 ext{ZnSO}_4 + 2 ext{K}_4[ ext{Fe}( ext{CN})_6] ightarrow ext{K}_2 ext{Zn}_3[ ext{Fe}( ext{CN})_6]_2 + 2 ext{K}_2 ext{SO}_4\]This tells us that 2 moles of \( ext{K}_4[ ext{Fe}( ext{CN})_6]\) react with 3 moles of \( ext{ZnSO}_4\).

Calculate Moles of ZnSO4

Next, we calculate the moles of \( ext{ZnSO}_4\) initially present. We are given 60 mL of a 0.01 M solution:\[ ext{moles of ZnSO}_4 = 0.01 ext{ M} \times 0.060 ext{ L} = 0.0006 ext{ mol}\]

Use Stoichiometry to Find Moles of K4[Fe(CN)6]

According to the balanced equation, 3 moles of \( ext{ZnSO}_4\) react with 2 moles of \( ext{K}_4[ ext{Fe}( ext{CN})_6]\). Therefore, for 0.0006 moles of \( ext{ZnSO}_4\) we need:\[ ext{moles of K}_4[ ext{Fe}( ext{CN})_6] = \frac{2}{3} \times 0.0006 = 0.0004 ext{ mol}\]

Calculate Volume of K4[Fe(CN)6] Solution Required

We now find the volume of 0.05 M \( ext{K}_4[ ext{Fe}( ext{CN})_6]\) solution that contains 0.0004 moles:\[ ext{Volume (L)} = \frac{0.0004 ext{ mol}}{0.05 ext{ M}} = 0.008 ext{ L}\]Convert to milliliters:\[0.008 ext{ L} = 8 ext{ mL}\]

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemistry. It helps us understand the precise amount of reactants needed to get a desired amount of product in a chemical reaction.
By using the balanced chemical equation, we can find out how much of each substance is needed or produced. In simple terms, it's about understanding the relationships between reactants and products based on their moles or molecules.

Here's why stoichiometry is important:

• It ensures that no reactant is wasted.
• It helps predict the number of products formed. This can be very important in laboratory and industrial applications.
• It's essential for mixing reactants in the correct proportions, which is crucial for safety and efficiency.

This comes into play when we relate the moles of one chemical to another using their coefficients in a balanced chemical equation.
For instance, in the given problem, we used stoichiometry to determine that 2 moles of \( \text{K}_4[ \text{Fe}( \text{CN})_6] \) were required for every 3 moles of \( \text{ZnSO}_4 \). By using stoichiometry, we ensure that all \( \text{ZnSO}_4 \) reacts without any surplus.

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding any chemical reaction. It shows the reactants on the left and the products on the right, with coefficients indicating the number of moles involved.
Balancing a chemical equation means ensuring the number of each type of atom on the reactants side is equal to the number on the products side.

Why is balancing the chemical equation important?

• It ensures the law of conservation of mass is followed, meaning matter is neither created nor destroyed.
• It allows us to use stoichiometry properly by providing the right ratios between substances.

In this specific exercise, the balanced equation is:\[3 \text{ZnSO}_4 + 2 \text{K}_4[ \text{Fe}( \text{CN})_6] \rightarrow \text{K}_2 \text{Zn}_3[ \text{Fe}( \text{CN})_6]_2 + 2 \text{K}_2 \text{SO}_4\]This equation shows that 3 moles of \( \text{ZnSO}_4 \) react with 2 moles of \( \text{K}_4[ \text{Fe}( \text{CN})_6] \). By understanding this balance, we can accurately predict how much product forms and how much of each reactant is needed.

Moles Calculation

Understanding moles is key to understanding chemistry, as it allows chemists to quantify atoms and molecules in a measurable unit. One mole is equal to Avogadro's number, \(6.022 \times 10^{23}\), representing the number of particles or molecules.
To calculate moles, you can use the formula: \[\text{moles} = \text{concentration (Molarity)} \times \text{Volume (in Liters)}\]

Here's why moles calculation is important:

• It allows chemists to take measurable amounts and translate them to molecular quantities.
• It is crucial for determining the precise amount of a reactant to use in a chemical reaction.
• It helps determine the volume needed to achieve a certain concentration.

In the problem provided, we calculated the moles of \( \text{ZnSO}_4 \) by multiplying its molarity by the volume in liters, which gave us \(0.0006\) moles.
Then, using stoichiometry, we found out the required moles of \( \text{K}_4[ \text{Fe}( \text{CN})_6] \) and finally converted this into the necessary volume of solution needed for titration.