From your Dilution Data in the procedure 3 and 4, calculate the initial concentration of Fe3+ (aq) ion for tubes 2-5.
1. Add 5.00ml of 1.00 x10^-3 M KSCN to all tubes.
2. To test tube 1 add 5 ml of 0.200M Fe(NO3)3.
3. Measure 8.00ml of 0.200M Fe(NO3)3 into 25.0ml cylinder. Fill to the 20ml mark with water. Add 5.00ml solution to tube 2. Save the remainder.
4. Pour 8.0ml solution in the beaker into 25.0ml cylinder. Fill to 20ml with water. Pour 5 ml solution into tube 3.
5. Repeat steps 4 and 5.
I don't understand the dilution to figure out the initial concentraions of Fe3+ of each one. Plz help.
2. You had 5 mL in the tube initially (of KSCN but that's still 5 mL). You add 5 mL of 0.2 M Fe(NO3)3 so the iron nitrate has been diluted from to 10 mL; therefore, its concentration has been halved. That is 0.2 M x (5/10) = ?
I'm assuming that the last sentence about adding 5.00 mL "solution" does NOT contain iron.
3. For this one you start with 0.2 M, and you go from 8 to 25. New concentration is 0.2 M x (8/25) = ?
4. I don't know what was in the beaker.
So far in 1,2,3 you've only talked about tubes and cylinders.
To calculate the initial concentration of Fe3+ in tubes 2-5, we need to understand the dilution process in steps 3-5 of the procedure. Let's break it down:
Step 3:
- Measure 8.00 ml of 0.200 M Fe(NO3)3 into a 25.0 ml cylinder.
- Fill the cylinder to the 20 ml mark with water.
- Add 5.00 ml of this diluted solution to tube 2.
- Save the remaining diluted solution for later use.
In this step, we take 8.00 ml of the original 0.200 M Fe(NO3)3 solution and dilute it to a total volume of 20 ml by adding water. We then take 5.00 ml of this diluted solution and add it to tube 2. So, the concentration of Fe3+ in tube 2 is equal to the concentration of the diluted solution.
Step 4:
- Pour 8.0 ml of the solution in the beaker into a 25.0 ml cylinder.
- Fill the cylinder to the 20 ml mark with water.
- Pour 5 ml of this diluted solution into tube 3.
In this step, we take 8.0 ml of the remaining diluted solution from Step 3 and dilute it to a total volume of 20 ml by adding water. We then take 5 ml of this further diluted solution and add it to tube 3. So, the concentration of Fe3+ in tube 3 is equal to the concentration of this additional diluted solution.
Step 5:
- Repeat Step 4.
This step involves the same process as Step 4, where we take 8.0 ml of the remaining diluted solution and dilute it to a total volume of 20 ml. We then take 5 ml of this additional diluted solution and add it to tube 4.
To calculate the initial concentration of Fe3+ in each tube, we need to know the original concentration of the Fe(NO3)3 solution used. In this case, it is given as 0.200 M.
Now, let's calculate the initial concentrations of Fe3+ in tubes 2-5.
Tube 2:
The concentration of Fe3+ in tube 2 is equal to the concentration of the diluted solution from Step 3, which is the same as the original concentration of Fe(NO3)3, i.e., 0.200 M.
Tube 3:
The concentration of Fe3+ in tube 3 is equal to the concentration of the additional diluted solution from Step 4. We started with 8.0 ml of the diluted solution from Step 3 and diluted it to a total volume of 20 ml, so the final volume is (8.0/20) times the initial concentration. Therefore, the concentration of Fe3+ in tube 3 is (8.0/20) * 0.200 M.
Tube 4:
Similar to tube 3, we started with the remaining diluted solution from Step 4 and diluted it to a total volume of 20 ml. So the concentration of Fe3+ in tube 4 is (8.0/20) * [(8.0/20) * 0.200] M.
Tube 5:
Again, we repeat the same dilution process as in Step 4 and Tube 4. So the concentration of Fe3+ in tube 5 is (8.0/20) * [(8.0/20) * [(8.0/20) * 0.200]] M.
Now, you can use these calculations to determine the initial concentration of Fe3+ in tubes 2-5 by plugging in the values for the dilution ratios and the original concentration of Fe(NO3)3.