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Sagot :
Let's go through the steps to solve the problem systematically.
### Step 1: Given Information
1. Initial Concentrations:
- Iodide ([tex]\(I^-\)[/tex]): [tex]\(100.0 \text{ mM}\)[/tex]
- Thiosulfate ([tex]\(S_2O_3^{2-}\)[/tex]): [tex]\(31.5 \text{ mM}\)[/tex]
- Hydrogen Peroxide ([tex]\(H_2O_2\)[/tex]): [tex]\(176.5 \text{ mM}\)[/tex]
2. Volumes Used:
- Iodide solution: [tex]\(10.0 \text{ mL}\)[/tex]
- Thiosulfate solution: [tex]\(1.0 \text{ mL}\)[/tex]
- Hydrogen peroxide solution: [tex]\(2.5 \text{ mL}\)[/tex]
- DI water: To be calculated to fill up to [tex]\(20.0 \text{ mL}\)[/tex]
3. Time:
- [tex]\(7.5 \text{ s}\)[/tex]
4. Observations:
- The solution turns blue.
### Step 2: Total Volume Calculation
We need the total volume of the reaction mixture.
Given:
- Volume of iodide solution: [tex]\(10.0 \text{ mL}\)[/tex]
- Volume of thiosulfate solution: [tex]\(1.0 \text{ mL}\)[/tex]
- Volume of hydrogen peroxide solution: [tex]\(2.5 \text{ mL}\)[/tex]
- Volume of DI water: [tex]\(\text{7.5 mL (since we need to fill to 20 mL total volume)}\)[/tex]
Total volume:
[tex]\[ \text{Total Volume} = 10.0 \text{ mL} + 1.0 \text{ mL} + 2.5 \text{ mL} + 7.5 \text{ mL} = 21.0 \text{ mL} \][/tex]
But the observation states the final volume should be 20 mL.
Refine DI water volume:
[tex]\[ \text{DI water volume} = 20.0 \text{ mL} - (10.0 \text{ mL} + 1.0 \text{ mL} + 2.5 \text{ mL}) = 6.5 \text{ mL} \][/tex]
### Step 3: Calculate Initial Concentrations in Reaction Mixture
To find the initial concentrations of iodide, thiosulfate, and hydrogen peroxide in the 20 mL reaction mixture, we use the dilution formula:
[tex]\[ C_{\text{initial}} = \frac{ C_{\text{stock}} \times V_{\text{stock}} } { V_{\text{total}} } \][/tex]
1. Initial Iodide Concentration:
[tex]\[ C_{I^-} = \frac{100.0 \text{ mM} \times 10.0 \text{ mL}}{20.0 \text{ mL}} = 47.6 \text{ mM} \][/tex]
2. Initial Thiosulfate Concentration:
[tex]\[ C_{S_2O_3^{2-}} = \frac{31.5 \text{ mM} \times 1.0 \text{ mL}}{20.0 \text{ mL}} = 1.575 \text{ mM} \][/tex]
3. Initial Hydrogen Peroxide Concentration:
[tex]\[ C_{H_2O_2} = \frac{176.5 \text{ mM} \times 2.5 \text{ mL}}{20.0 \text{ mL}} = 22.0625 \text{ mM} \][/tex]
### Step 4: Observations and Initial Rate
1. Observations:
- Volume of iodide solution ([tex]\(I^-\)[/tex]) used: [tex]\(10.0 \text{ mL}\)[/tex]
- Volume of thiosulfate solution ([tex]\(S_2O_3^{2-}\)[/tex]) used: [tex]\(1.0 \text{ mL}\)[/tex]
- Volume of hydrogen peroxide solution ([tex]\(H_2O_2\)[/tex]) used: [tex]\(2.5 \text{ mL}\)[/tex]
- Volume of DI water: [tex]\(6.5 \text{ mL}\)[/tex]
- The solution turns blue.
2. Initial Rate:
- From the given data, we know it is [tex]\(1.5 \text{ m/M/s}\)[/tex].
### Summary of Results
- Initial Concentration of Iodide ([tex]\(I^-\)[/tex]): [tex]\(47.6 \text{ mM}\)[/tex]
- Initial Concentration of Thiosulfate ([tex]\(S_2O_3^{2-}\)[/tex]): [tex]\(1.575 \text{ mM}\)[/tex]
- Initial Concentration of Hydrogen Peroxide ([tex]\(H_2O_2\)[/tex]): [tex]\(22.0625 \text{ mM}\)[/tex]
- Time: [tex]\(7.5 \text{ s}\)[/tex]
- Initial Rate: [tex]\(1.5 \text{ m/M/s}\)[/tex]
These values represent the conditions under which the liquid turns blue within the observation time.
### Step 1: Given Information
1. Initial Concentrations:
- Iodide ([tex]\(I^-\)[/tex]): [tex]\(100.0 \text{ mM}\)[/tex]
- Thiosulfate ([tex]\(S_2O_3^{2-}\)[/tex]): [tex]\(31.5 \text{ mM}\)[/tex]
- Hydrogen Peroxide ([tex]\(H_2O_2\)[/tex]): [tex]\(176.5 \text{ mM}\)[/tex]
2. Volumes Used:
- Iodide solution: [tex]\(10.0 \text{ mL}\)[/tex]
- Thiosulfate solution: [tex]\(1.0 \text{ mL}\)[/tex]
- Hydrogen peroxide solution: [tex]\(2.5 \text{ mL}\)[/tex]
- DI water: To be calculated to fill up to [tex]\(20.0 \text{ mL}\)[/tex]
3. Time:
- [tex]\(7.5 \text{ s}\)[/tex]
4. Observations:
- The solution turns blue.
### Step 2: Total Volume Calculation
We need the total volume of the reaction mixture.
Given:
- Volume of iodide solution: [tex]\(10.0 \text{ mL}\)[/tex]
- Volume of thiosulfate solution: [tex]\(1.0 \text{ mL}\)[/tex]
- Volume of hydrogen peroxide solution: [tex]\(2.5 \text{ mL}\)[/tex]
- Volume of DI water: [tex]\(\text{7.5 mL (since we need to fill to 20 mL total volume)}\)[/tex]
Total volume:
[tex]\[ \text{Total Volume} = 10.0 \text{ mL} + 1.0 \text{ mL} + 2.5 \text{ mL} + 7.5 \text{ mL} = 21.0 \text{ mL} \][/tex]
But the observation states the final volume should be 20 mL.
Refine DI water volume:
[tex]\[ \text{DI water volume} = 20.0 \text{ mL} - (10.0 \text{ mL} + 1.0 \text{ mL} + 2.5 \text{ mL}) = 6.5 \text{ mL} \][/tex]
### Step 3: Calculate Initial Concentrations in Reaction Mixture
To find the initial concentrations of iodide, thiosulfate, and hydrogen peroxide in the 20 mL reaction mixture, we use the dilution formula:
[tex]\[ C_{\text{initial}} = \frac{ C_{\text{stock}} \times V_{\text{stock}} } { V_{\text{total}} } \][/tex]
1. Initial Iodide Concentration:
[tex]\[ C_{I^-} = \frac{100.0 \text{ mM} \times 10.0 \text{ mL}}{20.0 \text{ mL}} = 47.6 \text{ mM} \][/tex]
2. Initial Thiosulfate Concentration:
[tex]\[ C_{S_2O_3^{2-}} = \frac{31.5 \text{ mM} \times 1.0 \text{ mL}}{20.0 \text{ mL}} = 1.575 \text{ mM} \][/tex]
3. Initial Hydrogen Peroxide Concentration:
[tex]\[ C_{H_2O_2} = \frac{176.5 \text{ mM} \times 2.5 \text{ mL}}{20.0 \text{ mL}} = 22.0625 \text{ mM} \][/tex]
### Step 4: Observations and Initial Rate
1. Observations:
- Volume of iodide solution ([tex]\(I^-\)[/tex]) used: [tex]\(10.0 \text{ mL}\)[/tex]
- Volume of thiosulfate solution ([tex]\(S_2O_3^{2-}\)[/tex]) used: [tex]\(1.0 \text{ mL}\)[/tex]
- Volume of hydrogen peroxide solution ([tex]\(H_2O_2\)[/tex]) used: [tex]\(2.5 \text{ mL}\)[/tex]
- Volume of DI water: [tex]\(6.5 \text{ mL}\)[/tex]
- The solution turns blue.
2. Initial Rate:
- From the given data, we know it is [tex]\(1.5 \text{ m/M/s}\)[/tex].
### Summary of Results
- Initial Concentration of Iodide ([tex]\(I^-\)[/tex]): [tex]\(47.6 \text{ mM}\)[/tex]
- Initial Concentration of Thiosulfate ([tex]\(S_2O_3^{2-}\)[/tex]): [tex]\(1.575 \text{ mM}\)[/tex]
- Initial Concentration of Hydrogen Peroxide ([tex]\(H_2O_2\)[/tex]): [tex]\(22.0625 \text{ mM}\)[/tex]
- Time: [tex]\(7.5 \text{ s}\)[/tex]
- Initial Rate: [tex]\(1.5 \text{ m/M/s}\)[/tex]
These values represent the conditions under which the liquid turns blue within the observation time.
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