Hg2+ + 2Cl- → HgCl

Mercuric nitrate was standardized by titrating a solution containing 147.6 mg of NaCl(FM 58.44), which required 28.06 ml of Hg(No3)2 solution.

Part a: Find the molarity of Hg(No3)2.

I already calculated the molarity to be 0.04500 M

Part b: When the same Hg(No3)2 solution was used to titrate 2.00 mL of urine, 22.83 mL was required. What is theconcentration of Cl- (mg/mL) in urine.

I know the answer for part b is 36.42 mg/ml, but I'm not sure exactly how to get there...

Thanks.

1)Calculate the molarity for the mercuric nitrate solution. =0.04500M

2)Use M1V1=M2V2, to find urine concentration. (0.045mol/L)(0.02283L)=(M2)(0.002L), M2=0.513675mol/L

3)Find concentration of Cl using titration equation molecular ratios and molecular mass of Cl.
(0.513675molUrine/L)*(1molHgCl2/1molUrine)*(2molCl/1molHgCl2)*(35.453gCl/1molCl)*(1L/1000mL)*(1000mg/1g)=36.42mg/mL

Asuna, How did you calculate the Molarity of in part A?

To find the concentration of Cl- in urine, we need to use the stoichiometry of the reaction between Hg(No3)2 and Cl-. We know that one mole of Hg(No3)2 reacts with two moles of Cl- to form HgCl.

First, let's calculate the number of moles of Hg(No3)2 used in the titration:

Moles of Hg(No3)2 = Molarity of Hg(No3)2 solution × Volume of Hg(No3)2 solution (in liters)
= 0.04500 M × (28.06 mL / 1000 mL/ L)
= 0.00126 moles of Hg(No3)2

Now, we apply the stoichiometry of the reaction to calculate the number of moles of Cl- present in the titrated amount of Hg(No3)2:

Moles of Cl- = 2 × Moles of Hg(No3)2
= 2 × 0.00126 moles
= 0.00252 moles of Cl-

Next, let's calculate the concentration of Cl- in the urine sample:

Concentration of Cl- (mg/mL) = (Moles of Cl- × Molar mass of Cl-) / Volume of urine sample (in mL)
= (0.00252 moles × 35.45 g/mol) / 2.00 mL
= 0.08925 g/mL
= 89.25 mg/mL

Therefore, the concentration of Cl- in the urine sample is 89.25 mg/mL.

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