This lab was a mercurimetric determination of blood chloride. I was able to calc

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This lab was a mercurimetric determination of blood chloride. I was able to calculate the conc. of the standardized Hg(NO3)2: 0.00905M. Now I have to calculate the Cl- content of the sample--first in mmol/L then in meq/L. [In one of my titrations I added 5.557 mL of Hg(NO3)2.] MERCURIMETRIC DETERMINATION oF BLOOD CHLORIDE Reference: see Day & Underwood, 6th Ed., pp. 220-221. Equipment: Chemicals 50 mL vol. flask HNOy 100 mL vol. flask Naci 250 mL vol. flask H2SO4 100 ml. beaker Na2WOa 10 mL pipet s-diphenyl carbazone (indicator) 10 mL buret 5 mL serological pipet 3-50 mL erlenmeyer flasks 10 mL serological pipet Preparation and Standardization of 0.009 E Hg(NO3)2: Procedure Explanation Precautions 1. Prep. of 0.009 F Hg(NO) [Lab This is ca. 0.009 F Hg(NO) Since the Solution returned to you by lab instr. will provide in your clean 100 orig. employed solid Hg(NO)2 is very instructor may be concentrated, i e., you mL volumetric flask.) hygroscopic (cannot accurately weigh), may have to dilute to the mark before we will have to standardize with Naci. use. Consult with your lab instructor *2. Independently prepare a std. Cl soln. by drying (1 hr.), cooling (1/2 Since the F W. of Naci w 58.443, ss D hr.), and accurately weighing 0.14 0.58443 g/L or 0.14611 g/250 mL would (t0.0001) g Naci into 250 mL vol. correspond exactly to 0.01000 F flask. Dilute to mark with H20. Mix. 3. Pipet 10.00 ml, 0.01 F NaCl into 50 be in one crude sample should first be run to mL erlenmeyer flask. Add 10 drops Steps 3 & 4 should performed triplicate. Results should show rel. dev. obtain an approximate end point indicator. of ca. 5 p.p.t. estimate for the 3 samples D-1

Explanation / Answer

Hg^2+ + 2Cl^- ==> HgCl2

1mol Hg(NO3)2 reacts with 2 mol Cl-

moles of Hg(NO3)2 added = 0.00905 M * 0.00557 L = 5.04*10^-5 moles

Moles of Cl- present = 2 * 5.04*10^-5 moles =1.0082*10^-4 moles

1.0082*10^-4 moles Cl^- is present in 5mL sample.

Concentration of Cl- in 1000mL = 1.0082*10^-4 *1000mL = 0.010082 mol/L = 0.010082 *10^-3 mmol/L

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