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### a. Balanced Equation
First, write the balanced chemical equation for the reaction between oxalic acid (H₂C₂O₄) and sodium hydroxide (NaOH). Especially noteworthy is that oxalic acid is diprotic, meaning it has two acidic protons:
[tex]\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \][/tex]
### b. Moles of Oxalic Acid
The moles of oxalic acid present in the sample can be calculated using its concentration and volume. The formula for moles is:
[tex]\[ \text{Moles of oxalic acid} = \text{Concentration} \times \text{Volume} \][/tex]
Given:
- Volume of oxalic acid solution, [tex]\( V = 30.0 \, \text{mL} = 0.0300 \, \text{L} \)[/tex] (convert mL to L by dividing by 1000)
- Concentration of oxalic acid, [tex]\( C = 0.2488 \, M \)[/tex]
Now, calculate the moles of oxalic acid:
[tex]\[ \text{Moles of oxalic acid} = 0.2488 \, \text{M} \times 0.0300 \, \text{L} = 0.007464 \, \text{moles} \][/tex]
### c. Moles of NaOH Needed
From the balanced equation, 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH needed are twice the moles of oxalic acid:
[tex]\[ \text{Moles of NaOH} = 2 \times \text{Moles of oxalic acid} \][/tex]
Using the moles of oxalic acid calculated earlier:
[tex]\[ \text{Moles of NaOH} = 2 \times 0.007464 \, \text{moles} = 0.014928 \, \text{moles} \][/tex]
### d. Molar Concentration of NaOH
To find the molar concentration of the NaOH solution, we use the formula for concentration, which is the number of moles divided by the volume in liters:
[tex]\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH solution in liters}} \][/tex]
Given:
- Volume of NaOH solution, [tex]\( V = 31.92 \, \text{mL} = 0.03192 \, \text{L} \)[/tex]
Using the moles of NaOH calculated earlier:
[tex]\[ \text{Concentration of NaOH} = \frac{0.014928 \, \text{moles}}{0.03192 \, \text{L}} \approx 0.4677 \, \text{M} \][/tex]
### Summary:
- a. Balanced Equation: [tex]\(\text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O}\)[/tex]
- b. Moles of oxalic acid: [tex]\(0.007464 \, \text{moles}\)[/tex]
- c. Moles of NaOH needed: [tex]\(0.014928 \, \text{moles}\)[/tex]
- d. Molar concentration of NaOH: [tex]\(0.4677 \, \text{M}\)[/tex]
### a. Balanced Equation
First, write the balanced chemical equation for the reaction between oxalic acid (H₂C₂O₄) and sodium hydroxide (NaOH). Especially noteworthy is that oxalic acid is diprotic, meaning it has two acidic protons:
[tex]\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \][/tex]
### b. Moles of Oxalic Acid
The moles of oxalic acid present in the sample can be calculated using its concentration and volume. The formula for moles is:
[tex]\[ \text{Moles of oxalic acid} = \text{Concentration} \times \text{Volume} \][/tex]
Given:
- Volume of oxalic acid solution, [tex]\( V = 30.0 \, \text{mL} = 0.0300 \, \text{L} \)[/tex] (convert mL to L by dividing by 1000)
- Concentration of oxalic acid, [tex]\( C = 0.2488 \, M \)[/tex]
Now, calculate the moles of oxalic acid:
[tex]\[ \text{Moles of oxalic acid} = 0.2488 \, \text{M} \times 0.0300 \, \text{L} = 0.007464 \, \text{moles} \][/tex]
### c. Moles of NaOH Needed
From the balanced equation, 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH needed are twice the moles of oxalic acid:
[tex]\[ \text{Moles of NaOH} = 2 \times \text{Moles of oxalic acid} \][/tex]
Using the moles of oxalic acid calculated earlier:
[tex]\[ \text{Moles of NaOH} = 2 \times 0.007464 \, \text{moles} = 0.014928 \, \text{moles} \][/tex]
### d. Molar Concentration of NaOH
To find the molar concentration of the NaOH solution, we use the formula for concentration, which is the number of moles divided by the volume in liters:
[tex]\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH solution in liters}} \][/tex]
Given:
- Volume of NaOH solution, [tex]\( V = 31.92 \, \text{mL} = 0.03192 \, \text{L} \)[/tex]
Using the moles of NaOH calculated earlier:
[tex]\[ \text{Concentration of NaOH} = \frac{0.014928 \, \text{moles}}{0.03192 \, \text{L}} \approx 0.4677 \, \text{M} \][/tex]
### Summary:
- a. Balanced Equation: [tex]\(\text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O}\)[/tex]
- b. Moles of oxalic acid: [tex]\(0.007464 \, \text{moles}\)[/tex]
- c. Moles of NaOH needed: [tex]\(0.014928 \, \text{moles}\)[/tex]
- d. Molar concentration of NaOH: [tex]\(0.4677 \, \text{M}\)[/tex]
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