The chloride content of blood serum, cerebrospinal fluid or urine can be measure

ID: 848572 • Letter: T

Question

The chloride content of blood serum, cerebrospinal fluid or urine can be measured

by titration of the chloride with mercuric ion: Hg2+ + 2Cl---> HgCl2(aq). When

the reaction is complete, excess Hg2+ reacts with the indicator,

diphenylcarbazone, which forms a violet-blue color.

(a) Mercuric nitrate was standardized by titrating a solution containing 147.6

mg of NaCl, which required 28.06 mL of Hg(NO3)

2 solution. Find the

molarity of the Hg(NO3)

(b) When this same Hg(NO3)

2 solution was used to titrate 2.000 mL of urine,

22.83 mL was required. Find the concentration of Cl-

(mg/mL) in the

urine.

the answer is 36.42mg/mL for part 2. I never get that answer!

(a) Mass of NaCl taken = 147.6 *10-3 g

So, moles of NaCl in this mass = 147.6*10-3/MW = 147.6*10-3/58.5 = 2.52*10-3 moles

According to reaction stoichiometry, 2 moles Cl- react with 1 mole Hg2+

Thus 2.52*10-3 moles of Cl- will react with : (2.52*10-3)/2 = 1.26*10-3 moles Hg2+

Thus, molarity of the solution = moles of Hg2+/volume of solution in liters = (1.26*10-3)/(28.06*10-3) = 0.045 M

(b) Volume of solution needed = 23.83 ml

Thus, moles of Hg2+ needed = volume needed in liters*molarity = 23.83*10-3*0.045 = 1.07*10-3 moles

Thus, moles of Cl- in urine = 2*1.07*10-3 = 2.14*10-3

Thus, mass of Cl- in urine = 2.14*10-3*35.5 = 75.97 mg

So, conc of Cl- in urine = mass of Cl- in urine / volume of urine taken = 75.97/2 = 37.98 mg/ml

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