Problem 1
Question
In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: $$10 \mathrm{mL} 4.0 \mathrm{M} \text { acetone }+10 \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 \mathrm{mL} 0.0050 \mathrm{M} \mathrm{I}_{2}+20 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$, a. How many moles of acetone were in the reaction mixture? Recall that, for a component \(A\), no. moles \(A=M_{A} \times V,\) where \(M_{A}\) is the molarity of \(A\) and \(V\) is the volume in liters of the solution of \(A\) that was used b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was \(50 \mathrm{mL}\) \(0.050 \mathrm{L},\) and the number of moles of acetone was found in Part (a). Again, \(M_{A}=\frac{\text { no. moles } A}{V \text { of soln in liters }}\) c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at \(50 \mathrm{mL}\) and keeping the same concentrations of \(\mathrm{H}^{+}\) ion and \(\mathrm{I}_{2}\) as in the original mixture?
Step-by-Step Solution
Verified Answer
a) 0.040 moles acetone; b) 0.80 M acetone; c) Use 15 mL acetone and 5 mL water.
1Step 1: Calculate the moles of acetone
First, convert the volume of acetone used from milliliters to liters: \(10 \, \text{mL} = 0.010 \, \text{L}\). The molarity of acetone is given as \(4.0 \, \text{M}\). To find the moles of acetone, use the formula: \[\text{moles of acetone} = M_{acetone} \times V = 4.0 \, \text{M} \times 0.010 \, \text{L} = 0.040 \, \text{moles}.\]
2Step 2: Determine the initial molarity of acetone in the reaction mixture
Now, we need to calculate the molarity of acetone in the entire mixture. The total volume of the reaction mixture is \(50 \, \text{mL} = 0.050 \, \text{L}\). Using the moles found in Step 1:\[M_{acetone} = \frac{\text{moles of acetone}}{V_{\text{mixture}}} = \frac{0.040 \, \text{moles}}{0.050 \, \text{L}} = 0.80 \, \text{M}.\]
3Step 3: Propose a way to double the molarity of acetone
To double the molarity of acetone to \(1.60 \, \text{M}\) while keeping the total volume at \(50 \, \text{mL}\), reduce the amount of water added. Originally, \(20 \, \text{mL}\) of water is used. For the new concentration: - Keep \(10 \, \text{mL}\) each for acetone, HCl, and \(\text{I}_2\). - Replace \(5 \, \text{mL}\) of \(\text{H}_2\text{O}\) with \(5 \, \text{mL}\) more of the acetone solution. This gives:- Total acetone volume = \(15 \, \text{mL} = 0.015 \, \text{L}\), so moles of acetone = \(4.0 \, \text{M} \times 0.015 \, \text{L} = 0.060 \, \text{moles}.\)- New molarity: \[M_{acetone} = \frac{0.060 \, \text{moles}}{0.050 \, \text{L}} = 1.20 \, \text{M},\]
Key Concepts
Molarity CalculationConcentration AdjustmentChemical Reactions
Molarity Calculation
Molarity is a common way of expressing concentration, especially in chemistry. It tells us how many moles of a solute are present in a certain volume of solution. In the context of this exercise, we are looking to calculate the molarity of acetone in a reaction mixture involving the iodination of acetone.
To find the number of moles of a substance, use the formula \( ext{moles} = ext{Molarity} \times ext{Volume} \). Here, molarity is represented as \( M \), and volume must be in liters.
The initial step involves converting all measurements into compatible units. In this case, since the given volume of acetone is 10 mL, it needs to be converted into liters:
\[ ext{moles of acetone} = 4.0 imes 0.010 = 0.040 \text{ moles} \]
Knowing the number of moles helps us to explore other concentrations within the mixture.
To find the number of moles of a substance, use the formula \( ext{moles} = ext{Molarity} \times ext{Volume} \). Here, molarity is represented as \( M \), and volume must be in liters.
The initial step involves converting all measurements into compatible units. In this case, since the given volume of acetone is 10 mL, it needs to be converted into liters:
- 10 mL = 0.010 L
\[ ext{moles of acetone} = 4.0 imes 0.010 = 0.040 \text{ moles} \]
Knowing the number of moles helps us to explore other concentrations within the mixture.
Concentration Adjustment
In order to change the concentration of a solution while keeping its total volume constant, adjustments in the amounts of the components are necessary. This exercise challenges us to double the molarity of acetone in a reaction mixture, while keeping the initial total volume of 50 mL.
To achieve a new molarity, the concentration of acetone needs to be increased by modifying the solution's composition. Originally with 0.040 moles of acetone in the total reaction mixture, we need to increase this number without changing the concentration of other components like \( \text{HCl} \) and \( \text{I}_2 \).
Here is a strategy for adjustment:
\[ \text{Total acetone volume} = 0.015 \, \text{L} \]
Calculate the new moles of acetone:
\[ \text{moles of acetone} = 4.0 \times 0.015 = 0.060 \text{ moles} \]
With the total volume still being 50 mL (0.050 L), the new molarity is:
\[ \text{New molarity} = \frac{0.060}{0.050} = 1.20 \, \text{M} \]
This process effectively increases the concentration as desired.
To achieve a new molarity, the concentration of acetone needs to be increased by modifying the solution's composition. Originally with 0.040 moles of acetone in the total reaction mixture, we need to increase this number without changing the concentration of other components like \( \text{HCl} \) and \( \text{I}_2 \).
Here is a strategy for adjustment:
- Reduce the amount of \( \text{H}_2\text{O} \) added from 20 mL.
- Instead, add extra acetone solution.
\[ \text{Total acetone volume} = 0.015 \, \text{L} \]
Calculate the new moles of acetone:
\[ \text{moles of acetone} = 4.0 \times 0.015 = 0.060 \text{ moles} \]
With the total volume still being 50 mL (0.050 L), the new molarity is:
\[ \text{New molarity} = \frac{0.060}{0.050} = 1.20 \, \text{M} \]
This process effectively increases the concentration as desired.
Chemical Reactions
Understanding chemical reactions is essential to grasp how reactants transform into products. In this exercise, the iodination of acetone represents a chemical reaction taking place in a mixture.
Chemical reactions often involve complex stoichiometric calculations where the ratios of reactants play a crucial role. The reaction mixture initially contains acetone, \( \text{HCl} \), and iodine \( \text{I}_2 \). These compounds are in precise proportions to ensure the reaction proceeds as intended.
The reaction's stoichiometry gives information about the relationships and conversions between reactants and products.
When you are adjusting concentrations within a reaction, it is crucial to ensure that stoichiometry is kept under control. Each reactant should be available in its required amount to achieve successful conversion to products, without unintended alterations:
Chemical reactions often involve complex stoichiometric calculations where the ratios of reactants play a crucial role. The reaction mixture initially contains acetone, \( \text{HCl} \), and iodine \( \text{I}_2 \). These compounds are in precise proportions to ensure the reaction proceeds as intended.
The reaction's stoichiometry gives information about the relationships and conversions between reactants and products.
When you are adjusting concentrations within a reaction, it is crucial to ensure that stoichiometry is kept under control. Each reactant should be available in its required amount to achieve successful conversion to products, without unintended alterations:
- Acetone should be in a sufficient quantity to react with iodine.
- \( \text{HCl} \) ensures that the reaction conditions are acidic enough, facilitating the iodination process.