A solution was prepared by adding 15mL of 0.129 M KI, 5mL of 0.1 M Na2S2O3, 1mL of starch solution, and 15mL of 0.125 M (NH4)2S2O8.
Calculate the concentration of (M) of I in the final mixture before the chemical reaction.
Get the moles of I from KI. then, get the volume by adding the volumes of all the chem solutions added.
To calculate the concentration of I in the final mixture before the chemical reaction, we need to determine the number of moles of I present and then divide by the total volume of the mixture.
Step 1: Calculate the number of moles of I in each solution added:
- For the KI solution:
- Volume (V1) = 15 mL
- Concentration (C1) = 0.129 M
- Moles of I (n1) = V1 * C1
- For the (NH4)2S2O8 solution:
- Volume (V2) = 15 mL
- Concentration (C2) = 0.125 M
- Moles of I (n2) = V2 * C2
Step 2: Determine the total moles of I in the mixture:
- Total moles of I = n1 + n2
Step 3: Calculate the total volume of the mixture:
- Total volume = sum of the volumes of all solutions added = V1 + V2 + V3 + V4
Step 4: Calculate the concentration of I in the final mixture:
- Concentration of I = Total moles of I / Total volume of the mixture
Now, let's calculate the values:
- For the KI solution:
- V1 = 15 mL
- C1 = 0.129 M
- n1 = V1 * C1 = 15 mL * (0.129 mol/L) = 1.935 mmol (converted to millimoles)
- For the (NH4)2S2O8 solution:
- V2 = 15 mL
- C2 = 0.125 M
- n2 = V2 * C2 = 15 mL * (0.125 mol/L) = 1.875 mmol
- Total moles of I = n1 + n2 = 1.935 mmol + 1.875 mmol = 3.81 mmol
- Total volume of the mixture = V1 + V2 + V3 + V4 = 15 mL + 15 mL + 5 mL + 1 mL = 36 mL
- Concentration of I = Total moles of I / Total volume of the mixture = 3.81 mmol / 36 mL = 0.106 M
Therefore, the concentration of I in the final mixture before the chemical reaction is approximately 0.106 M.