**Example Calculation of Percent Recovery in Drinking Water Analysis**A sample of drinking water was tested for Pb²⁺ and was found to have a Pb²⁺ concentration of 17.46 ppb. A 9.00 mL sample of the drinking water was spiked with 1.00 mL of a 2.29 ppb Pb²⁺ standard. Analysis of the spiked sample gave a concentration of 15.92 ppb Pb²⁺. Find the percent recovery of the spike.**% recovery = ______________ %****Instructions:**1. Calculate the theoretical concentration after spiking.2. Determine the concentration found experimentally in the spiked sample.3. Calculate percent recovery using the formula: \[ \text{Percent Recovery} = \left(\frac{\text{Found Concentration} - \text{Original Concentration}}{\text{Spiked Concentration}} \right) \times 100\% \]

Given the concentration of the unspiked sample, Cunspiked = 17.46 ppb Now 1.00 mL of 2.29 ppb Pb2+…

A sample of drinking water was tested for Pb+ and was found to have a Pb²* concentration of 17.46 ppb. A 9.00 mL sample of the drinking water was spiked with 1.00 mL of a 2.29 ppb Pb²+ standard. Analysis of the spiked sample gave a concentration of 15.92 ppb Pb2+. Find the percent recovery of the spike. % recovery =

A sample of drinking water was tested for Pb+ and was found to have a Pb²* concentration of 17.46 ppb. A 9.00 mL sample of the drinking water was spiked with 1.00 mL of a 2.29 ppb Pb²+ standard. Analysis of the spiked sample gave a concentration of 15.92 ppb Pb2+. Find the percent recovery of the spike. % recovery =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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