and and for disposal of chemicals, Exam problem A red standard solution of Ni(NH

Question

and and for disposal of chemicals, Exam problem A red standard solution of Ni(NHir(so40m 6H20 has been prepared with a concentration of 3.067 x 10- M. In solution, the compound completely breaks apart into its component parts, including the individual ion of interest Ni7 (aq). Calculate the ppm Niz (aa) for the standard solution. Show the value to the instructor Prepare three diluted solutions with Ni (aq concentrations of 4.50 ppm, 2.70 ppm and 1.35 ppm. Choose whatever glassware is best to achieve high accuracy for the dilution process. Each dilution volume should be no less than 10 mL (minimum to fill vial) and no more than 250 mL (to reduce chemical waste). Place each dilution in a clean vial. Determine the absorption of each diluted solution. The instructor will use the MicroLab interface to determine the absorbance and concentration of each diluted sample. calculate the molality and the by mass of Ni (aa) for the standard solution. Calculate the molarity of each diluted sample. Record the answers in the tables at the end of the exam.

Explanation / Answer

Preparation of Ni2+ standard solutions

We have,

3.067 x 10^-4 M Ni(NH4)2(SO4)2.6H2O solution

Ni2+ ppm in this solution = 3.067 x 10^-4 M x 58.69 g/mol x 1000

                                        = 18.00 ppm

To prepare dilute solutions from above stock solution,

4.50 ppm of 100 ml Ni2+ solution

Volume of stock solution to be taken = 4.50 ppm x 100 ml/18.00 ppm

                                                           = 25 ml

So 25 ml of 18 ppm solution diluted to 100 ml would give 4.50 ppm solution

molarity of solution = 4.50 mg/L/1000 x 58.69

                               = 7.67 x 10^-5 M

%mass Ni2+ = 0.0045 g x 100/1000 g

                      = 4.5 x 10^-4%

2.70 ppm of 100 ml Ni2+ solution

Volume of stock solution to be taken = 2.70 ppm x 100 ml/18.00 ppm

                                                           = 15 ml

So 15 ml of 18 ppm solution diluted to 100 ml would give 2.70 ppm solution

molarity of solution = 2.70 mg/L/1000 x 58.69

                               = 4.60 x 10^-5 M

%mass Ni2+ = 0.0027 g x 100/1000 g

                     = 2.7 x 10^-4%

1.35 ppm of 250 ml Ni2+ solution

Volume of stock solution to be taken = 1.35 ppm x 250 ml/18.00 ppm

                                                           = 18.75 ml

So 18.75 ml of 18 ppm solution diluted to 250 ml would give 1.35 ppm solution

molarity of solution = 1.35 mg/L/1000 x 58.69

                               = 2.30 x 10^-5 M

%mass Ni2+ = 0.00135 g x 100/1000 g

                     = 1.35 x 10^-4%

We would use 100 ml and 250 ml volumetric flasks. The stock solutions would be measured by using pipet of 50 ml and 25 ml each.

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