Question

1. You prepare a NaCl solution by dissolving making 250mL solution out of 0.1409 g of NaCl (MW = 58.443 g/mol). You use this solution to standardize a Hg(NO3)2 solution. The volume of Hg(NO3)2 needed to titrate 10 mL of NaCl is 4.017 mL what is the molarity of the a Hg(NO3)2 solution? Note: Hg2+ + 2 Cl → HgCl2 2. Using the Folin-Wu tungstic acid method you pipet 10 mL of a serum sample and precipitate the protein (Note: the final volume of filtrate is 50 mL). You pipet 5 mL of the filtrate and titrate it with 7.792 mL of 0.0093 M Hg(NO3)2 . Determine the meq/L of Cl' in the original serum sample. Note: Hg2+ + 2 Cl → HgCl2 3. What is the Folin-Wu method and what solutions are used?

Answer 1

Answer. 250 - 0.1409 mol of 0.1409 g of Nace = - Nach 580443 = 0.002411 mol of Nace. 0.002411 x 1000 M SO molarity of Nace solution prepared = 0.009644 M No. of moles of Nace in 10 ml solution = 0.0096 44 x 10 mol 1000 = 90640x10 mol 9.649x105 mol Nace = 9.648 x 10 mol ce... Let's assume molarity of Hg (NO3)2 is c (M)... Then, no. of moles of Hg (NO3)2 in 4.017 mL solution 2. . x4.017)mol = 0.0040176 mol

0.0040176 mol Hg (NO3)2 = 0.004017 c mol Hg at According to balanced equation - 1 mol of Hgat reacts with a moles of ce- -0.0040170 .. . (2x0-0040170) moles of us = 0.008034C moles of a So 0.008034c = 9.644810² a c= 0.01200 so molarity of Hg(NO3)2 solution - 10.01200M /