The % error in the extrapolated results is highly dependent on the error associa

Question

The % error in the extrapolated results is highly dependent on the error associated with the volume of the lead(II) nitrate solution used in each titration and the volume of potassium iodide solutions need to initiate precipitation in each trial. Given the data below, calculate the maximum % error associated with e of the typical measurements below. a. 100.00 mL of lead(II) nitrate solution measured using a 50.00 mL pipette with a tolerance of plusminus 0.05 mL b. 100.00 mL of lead(II) nitrate solution measured using a 100.00 mL volumetric flask with a tolerance of plusminus 0.02mL c 7.42 mL of potassium iodide solution titrated from a 50.00 mL buret with a tolerance of 10.02 mL Based on the calculations in question 1, which measurement introduces the greatest error? Can you suggest at least one possible modification to the procedure to improve the precision associated with individual titration points? Discuss how each of the following conditions might affect the volume of 0.200 M potassium iodide needed to reach the precipitation end point in a typical titration. a. The lead(II) nitrate solution was prepared in 0.05 M NaCl instead of deionized water. b. The temperature in the lab increased by 3 degree C.

Explanation / Answer

1a) We have to pipette out twice using 50.00 mL pipette with tolerance ±0.05mL for 100.00 mL. Therefore total tolerance in the lead (II) nitrate solution is ±0.1 mL.

Thus, % error = 0.1×100/100 (or 0.05×2×100/100) = 0.1

1b) Capacity of volumetric flask is 100.00±0.02 mL. Therefore while preparing the 100.00 mL lead (II) nitrate solution, we introduce 0.02 mL error

Thus, %error = 0.02×100/100 = 0.02

1c) Capacity of burette is 50.00±0.02 mL.

Therefore, tolerance per mL solution is 0.02/50 = 0.0004 mL and for 7.42 mL is 0.003 mL

Thus, % error = 0.003×100/7.42 = 0.04

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