A person with cystic fibrosis typically contains 60-200 mM Cl- in samples of the

Question

A person with cystic fibrosis typically contains 60-200 mM Cl- in samples of their perspiration. The chloride in a 1.0-mL aliquot of such a sample is to be measured by combining it with 1.0 mL of a reagent containing 0.0250 M AgNO3. Ksp = 1.8 times 10-10 What mass of AgCl would be expected to precipitate if the sample originally contained 200 mM Cl-? What mass of Cl- will remain in solution after this precipitation has occurred? What percent of Cl- in the original sample will have been precipitated during this analysis?

Explanation / Answer

200 mM = 0.2 M = 0.2 mole/L AgNO3 is limiting 0.2 mole/L * 0.001L = 0.0002 moles Cl - 0.025 mole/L * 0.001L = 0.000025 moles Ag + a) 0.001L * 0.025 mol/L * 143.32 g AgCl/mole = 0.003583 grams of AgCl b) (0.0002 moles - 0.000025 moles ) * 35.45 g/mole = 0.00620375 grams of Cl- c) 0.000025/0.0002 * 100% = 12.5%

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