Show all your work and explain each procedure.
A. 1.00 L of 2.00 M aqueous solution of Sodium Hydroxide (NaOH).
B. 90.0 mL of a 1.20 M aqueous solution of sodium oxalate (Na2 C2 O4) from a 2.00 M solution of Na2 C2 O4
The Correct Answer and Explanation is :
Let’s break down the problem and explain each procedure.
Part A: Calculating the Amount of Sodium Hydroxide (NaOH)
We are given a 2.00 M (molar) solution of sodium hydroxide (NaOH), and we need to find out how many moles of NaOH are in a 1.00 L (liter) solution.
Step 1: Use the molarity formula
The molarity formula is:
[
M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}}
]
Where:
- ( M ) is the molarity of the solution
- Moles of solute is the number of moles of NaOH
- Volume of solution is the amount of solution in liters
We know:
- ( M = 2.00 \, \text{M} )
- Volume of solution = 1.00 L
To find the moles of NaOH:
[
\text{Moles of NaOH} = M \times \text{Volume of solution (in L)}
]
[
\text{Moles of NaOH} = 2.00 \, \text{mol/L} \times 1.00 \, \text{L} = 2.00 \, \text{mol}
]
Thus, there are 2.00 moles of NaOH in 1.00 L of the 2.00 M solution.
Part B: Dilution of Sodium Oxalate (Na₂C₂O₄)
We are given a 2.00 M solution of sodium oxalate (Na₂C₂O₄), and we need to determine how to prepare 90.0 mL of a 1.20 M solution of sodium oxalate from this stock solution.
Step 1: Use the dilution equation
The dilution equation is:
[
M_1 V_1 = M_2 V_2
]
Where:
- ( M_1 ) is the molarity of the concentrated solution (2.00 M)
- ( V_1 ) is the volume of the concentrated solution that we need to take
- ( M_2 ) is the molarity of the diluted solution (1.20 M)
- ( V_2 ) is the volume of the diluted solution (90.0 mL or 0.090 L)
We need to solve for ( V_1 ) (the volume of the concentrated solution we will use):
[
V_1 = \frac{M_2 \times V_2}{M_1}
]
Substitute the known values:
[
V_1 = \frac{1.20 \, \text{M} \times 0.0900 \, \text{L}}{2.00 \, \text{M}} = \frac{0.108}{2.00} = 0.0540 \, \text{L}
]
So, ( V_1 = 0.0540 \, \text{L} ), or 54.0 mL.
Step 2: Prepare the diluted solution
To prepare 90.0 mL of 1.20 M sodium oxalate solution:
- Take 54.0 mL of the 2.00 M sodium oxalate solution.
- Add distilled water to make the total volume 90.0 mL.
This gives the required concentration of 1.20 M.
Summary of Results:
- In part A, there are 2.00 moles of NaOH in the 1.00 L solution of 2.00 M NaOH.
- In part B, to prepare 90.0 mL of a 1.20 M Na₂C₂O₄ solution from a 2.00 M stock solution, you need to use 54.0 mL of the stock solution and dilute it with water to a total volume of 90.0 mL.