Question

A) For an experiment you are doing in lab, you need 50.0mL of a 2.50 x 10-2 M solution of Ni(NO3)2. Describe how you would prepare this solution from solid nickel nitrate and water to make the most accurate and reproducible solution as possible. Include any necessary calculations and a description of each step you would follow in the lab. Indicate what items you would use from the “tool-box” shown on the next page. (Please limit your description toeightsentences or less.)

B) Your next experiment calls for 100.0 mL of a 2.50 x 10-5 M solution of nickel nitrate. How could you use the 2.50 x 10-2 M solution from your previous experiment to help you prepare this new solution such that it is as accurate and reproducible as possible? Include any necessary calculations and the items you would use from your “tool-box” for this stepin your explanation.(Please limit your description to eight sentences or less.)

C) After you make your second solution, one of your friends says they were about to make their 2.50 x 10-5 M solution from solid nickel nitrate and water instead of using up part of their first solution. Which of the two methods (your method or your friend’s proposed method) would be better to use if you wanted the molarity to be as accurate and reproducible as possible, and why? (Please limit your description to four sentences or less.) A) Making the 2.50 x 10-5 M solution from the 2.50 x 10-2 M solution B) Making the 2.50 x 10-5 M solution from solid nickel nitrate and water

Answer 1

Ans. #A. Given, molarity of Ni(NO₃)₂ = 2.50 x 10^-2 M = 0.0250 mol/ L

Required number of moles for 50.0 mL soln. = (0.0250 mol/ 1000 mL) x 0.050 mL

                                                            = 0.00125 mol

Required mass for 50.0 mL soln. = Required moles x Molar mass

                                                            = 0.00125 mol x (182.70328 g/ mol)

                                                            = 0.2283791 g

                                                            = 0.2284 g

                                                            = 228.4 mg

# Preparation: Accurately weigh 228.4 mg 0.228 g (depending on the availability of analytical or high precision balance available to you) of pure Ni(NO₃)₂ to a clean, class A 50-mL standard volumetric flask using suitable funnel.

Rinse the butter paper (or weighing paper) in funnel 2-3 times with small volume of deionized water to transfer traces of the crystals sticking on it into the flask.

Rinse the inner wall of funnel and its tip 2-3 times with deionized water to transfer all the traces of the compound into the flask.

Make the final volume upto the mark with deionized water. Mix well by inverting the stopper-fitted flask several times to get a homogenous solution.

It is you desired solution.

#B. Using       C1V1 (stock solution) = C2V2 (Diluted solution)

            Or, 2.50 x 10^-2 M x V1 = 2.50 x 10^-5 x 100.0 mL

            Or, V1 = (2.50 x 10^-5 x 100.0 mL) / (2.50 x 10^-2 M)

            Hence, V1 = 0.100 mL

Preparation: Transfer 0.100 mL (= 100.0 ug) of stock solution (#A soln.) using a suitable micropipette into a clean, class A 100-mL volumetric flask. Make the final volume upto the mark with deionized water. Mix well. It is the desired solution.

#C. Direct preparation:

Mass of Ni(NO₃)₂ required to prepare 100.0 mL of 2.50 x 10^-5 M solution =

                        (2.50 x 10^-5 M x 0.100 L) x (182.70328 g/ mol)

                        = 0.000457 g = 0.457 mg

Preparation: see #A preparation. Change mass and volume as needed.

# Comparison:

Most of the analytical balance in lab has readability of 0.0001 g or 0.1 mg with the last digit bearing uncertainty of +/- 1.

Weighing 0.457 mg in analytical balance may give different value every time, say one time 0.0004 g, next time 0.0005 or 0.0003 g and so on due to the uncertainty in last digit.

However, if #A solution is prepared using analytical balance, class A volumetric flasks and micropipette, a dilution to prepare #B gives a consistent, reproducible molarity because mass of solute is fixed all the times.

Therefore method #B gives reproducible results whereas #C may not.