Problem 52 - Chemistry and Chemical Reactivity

Question

Which of the following methods would you use to prepare \(300 .\) mL of \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (a) Add \(30.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to \(270 . \mathrm{mL}\) of water. (b) Dilute \(250 . \mathrm{mL}\) of \(0.600 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a volume of \(300 .\) mL.

Step-by-Step Solution

Answer

Use method (b): Dilute 250 mL of 0.600 M \(K_2Cr_2O_7\) to 300 mL.

Step 1: Determine Moles Needed

First, calculate the moles of \(K_2Cr_2O_7\) required for a \(0.500\, \mathrm{M}\) solution in \(300\, \mathrm{mL}\). Use the formula \(\text{Moles} = \text{Molarity} \times \text{Volume (L)}\). \[\text{Moles needed} = 0.500\, \mathrm{M} \times 0.300\, \mathrm{L} = 0.150\, \mathrm{moles}\]

Step 2: Evaluate Option (a)

Calculate the moles present in 30.0 mL of 1.50 M \(K_2Cr_2O_7\): \[\text{Moles from (a)} = 1.50\, \mathrm{M} \times 0.030\, \mathrm{L} = 0.045\, \mathrm{moles}\]Since 0.045 moles is less than the required 0.150 moles, Option (a) will not produce the desired concentration.

Step 3: Evaluate Option (b)

Calculate the moles present in 250 mL of 0.600 M \(K_2Cr_2O_7\):\[\text{Moles from (b)} = 0.600\, \mathrm{M} \times 0.250\, \mathrm{L} = 0.150\, \mathrm{moles}\]Since 0.150 moles matches the required amount from Step 1, diluting 250 mL to 300 mL maintains the intended molarity. Hence, Option (b) is valid.

Step 4: Confirm Final Decision

Option (b) correctly achieves the desired molarity when diluted to 300 mL because the number of moles (0.150) aligns with the requirement we calculated. The setup for Option (b) maintains the appropriate concentration.

Key Concepts

Molarity Calculation

Molarity is a way to express how concentrated a solution is. It tells us how many moles of a solute are in each liter of solution.
If you need to create a particular molar concentration, you first determine the number of moles needed using this formula:

• \( \text{Moles} = \text{Molarity} \times \text{Volume (L)} \)

In our exercise, we wanted a 0.500 M solution of \( K_2Cr_2O_7 \) in 300 mL of solution.

• This requires 0.150 moles (calculated by multiplying 0.500 M by 0.300 L).

Breaking down this calculation:

• First convert volume from milliliters to liters by dividing by 1000.
• Then multiply both molarity and volume in liters to get the moles.

Knowing how to calculate molarity helps in determining the right amounts of solute or solution needed for your chemistry experiments.

Dilution Process

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent.
This is common in chemistry labs and essential for accurately achieving the desired solution concentration.When diluting, the number of moles of the solute stays the same, but the volume of the solution changes.

• Use the formula \( M_1V_1 = M_2V_2 \) to relate initial and final states.

In our exercise, Option (b) involved diluting a 0.600 M solution from 250 mL to 300 mL to achieve a molarity of 0.500 M.

• This change in volume means the solution becomes less concentrated, matching the desired end molarity.

Diluting correctly ensures that experiments are conducted under the right conditions without using more solute than necessary.

Chemical Preparation Methods

Properly preparing a chemical solution is vital for accuracy and safety in a laboratory setting.
There are different methods you can use depending on the desired outcome.

• For high concentration needs, starting with a strong, known concentration and performing dilutions is effective.
• Option (a) in our exercise demonstrates a common mistake – using an incorrect ratio of solute to solvent, leading to the wrong concentration.
• Option (b) illustrates appropriate preparation through dilution to achieve precise molarity.

Choosing the correct preparation method connects to understanding your desired concentration and volume requirements.
Following safe and accurate preparation methods helps ensure successful experiments without wastage or error.