Sure! Let's break down the solution step-by-step.
### Step 1: Write the balanced redox reaction
The balanced redox reaction between permanganate ions ([tex]\(MnO_4^-\)[/tex]) and oxalate ions ([tex]\(C_2O_4^{2-}\)[/tex]) in an acidic solution is:
[tex]\[2 MnO_4^- + 5 C_2O_4^{2-} \rightarrow 2 Mn^{2+} + 10 CO_2\][/tex]
### Step 2: Determine the given quantities
- Molarity of [tex]\(MnO_4^-\)[/tex] [tex]\(= 0.200 \, \text{M}\)[/tex]
- Molarity of [tex]\(C_2O_4^{2-}\)[/tex] [tex]\(= 0.100 \, \text{M}\)[/tex]
- Volume of [tex]\(C_2O_4^{2-}\)[/tex] [tex]\(= 50.00 \, \text{mL}\)[/tex]
### Step 3: Convert the volume of [tex]\(C_2O_4^{2-}\)[/tex] from mL to L
[tex]\[V_{C_2O_4} = 50.00 \, \text{mL} \div 1000 = 0.050 \, \text{L}\][/tex]
### Step 4: Calculate the number of moles of [tex]\(C_2O_4^{2-}\)[/tex]
[tex]\[ \text{Moles of } C_2O_4^{2-} = \text{Molarity} \times \text{Volume} \][/tex]
[tex]\[ \text{Moles of } C_2O_4^{2-} = 0.100 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \][/tex]
### Step 5: Use the stoichiometry of the reaction to determine the moles of [tex]\(MnO_4^-\)[/tex] needed
The balanced equation shows that 2 moles of [tex]\(MnO_4^-\)[/tex] react with 5 moles of [tex]\(C_2O_4^{2-}\)[/tex]. Therefore, the moles of [tex]\(MnO_4^-\)[/tex] needed can be calculated using the ratio:
[tex]\[ \frac{2 \, \text{moles of } MnO_4^-}{5 \, \text{moles of } C_2O_4^{2-}} = \frac{x \, \text{moles of } MnO_4^-}{0.005 \, \text{moles of } C_2O_4^{2-}} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{2}{5} \times 0.005 \][/tex]
[tex]\[ x = 0.002 \, \text{moles of } MnO_4^- \][/tex]
### Step 6: Calculate the volume of [tex]\(0.200 \, \text{M} \, MnO_4^-\)[/tex] needed
[tex]\[ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} \][/tex]
[tex]\[ V_{MnO_4^-} = \frac{0.002 \, \text{moles}}{0.200 \, \text{M}} \][/tex]
[tex]\[ V_{MnO_4^-} = 0.010 \, \text{L} \][/tex]
Convert the volume from liters to milliliters:
[tex]\[ V_{MnO_4^-} = 0.010 \, \text{L} \times 1000 \, \text{mL/L} \][/tex]
[tex]\[ V_{MnO_4^-} = 10.00 \, \text{mL} \][/tex]
### Final Answer
The volume of [tex]\(0.200 \, \text{M} \, MnO_4^-\)[/tex] needed to react with 50.00 mL of [tex]\(0.100 \, \text{M} \, C_2O_4^{2-}\)[/tex] is 10.00 mL.
