Titration #1 Titration #2 Titration #3 The concentration of NaOH (M) 0.9554 0.9540 0.9551 Using the 3 concentrations, calculate the average concentration of the student's NaOH solution and calculate the ± error in parts per thousand (ppt). Show the complete calculation. Express your final answer in the form of average (with units) ± error in ppt.
Suppose a student performed a similar standardization titration experiment. Below is the calculated concentration of NaOH from each of their titrations.
| Titration #1 | Titration #2 | Titration #3 | |
|---|---|---|---|
| The concentration of NaOH (M) | 0.9554 | 0.9540 | 0.9551 |
Using the 3 concentrations, calculate the average concentration of the student's NaOH solution and calculate the ± error in parts per thousand (ppt). Show the complete calculation. Express your final answer in the form of average (with units) ± error in ppt.
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confused in this part why is the answer 0.0005666?
Therefore, average error/ deviation = d1 + d2 + d33= (0.0006+0.0008+0.0003)3= 0.0005666 ≈ 0.00057
error/ deviation in parts per thousand(ppt) = average error or deviationaverage concentration×1000 = 0.000570.9548×1000 = 0.596
So, the final answer is = 0.9548 M ± 0.596 ppt
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