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

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

Expert Solution

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

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

Ag⁺ + Cl^- -------> AgCl

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

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

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

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

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

= 3.31mg

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

= 30.28mg/L

queen_honey_blossom answered 2 years ago

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