100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ to titrate....

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Chemistry

100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ totitrate. Calculate the concentration of Cl- in ppm (Î¼g/mL or mg/L)for the river water.

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3.9 Ratings (482 Votes)

no of moles of Ag⁺Â Â = molarity * volume in L

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.01005*0.0093Â Â = 9.3465*10^-5 moles

Ag⁺ + Cl^- -------> AgCl

since Ag⁺ and Cl^- react in a 1:1 ratio

no of moles of Cl^-Â Â = no of moles of Ag⁺

no of moles of Cl^-Â Â = 9.3465*10^-5 moles

mass of Cl^-Â Â Â Â Â Â Â Â Â Â Â Â = no of moles * gram atomic mass

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =9.3465*10^-5 * 35.5 = 3.31*10^-3 g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3.31mg

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =3.31mg/0.1093L Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ total volume = 100+9.3 = 109.3ml =0.1093L]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 30.28mg/L

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