A chemist places 2.1041 g of Na2SO4 in a 100 mL volumetric flask and adds water

Question

A chemist places 2.1041 g of Na2SO4 in a 100 mL volumetric flask and adds water to the mark. She then pipets 10 mL of the resulting solution into a 50 mL volumetric flask and adds water to the mark and mixes to make a solution. She then pipets 15 mL of this new solution into a 25 mL volumetric flask and dilutes to the mark. Determine the molar concentration of sodium sulfate in the most dilute solution prepared.

A student obtains 100 mL of a 0.0003627 M solution of AgNO3 (silver nitrate). He labels this “solution #1”. He then pipets 5 mL of Solution #1 into a 50 mL volumetric flask and dilutes to the mark with water. He labels this “Solution #2.” He then pipets 10 mL of Solution #1 into a 250 mL volumetric flask and dilutes to the mark with water. This is “Solution #3”. Finally, the prepares “Solution #4” by pipetting 16 mL of solution #3 into a 50 mL volumetric flask and dilutes to the mark with water. Draw relational pictures to illustrate the relationship between the original solution and all of the diluted solutions that the student prepared. What is the concentration of silver nitrate in Solution #4?

A chemist obtains 500.0 mL of a solution containing an unknown concentration of calcium iodide, CaI2. He pipets 15 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pipets 10 mL of this diluted solution into a 100 mL volumetric flask and dilutes to the mark. He analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.0000067 M. (Note: in solution, calcium iodide breaks apart into one Ca2+ ion for every two I- ions, so a solution that is 1.0 M in CaI2 is 2.0 M in I-.) Determine the molar concentration of calcium iodide in the original solution.

Explanation / Answer

1) Molar mass of Na2SO4 = (2*22.9897 + 1*32.065 + 4*15.9994) g/mol = 142.042 g/mol.

Mole(s) of Na2SO4 corresponding to 2.1041 g = (moles of Na2SO4)/(molar mass of Na2SO4) = (2.1041 g)/(142.042 g/mol) = 0.0148 mole.

The said mole(s) of Na2SO4 were dissolved in 100 mL volumetric flask with water. The concentration of this solution (call this solution 1) = (moles of Na2SO4)/(volume of solution in L) = (0.0148 mole)/[(100 mL)*(1 L/1000 mL)] = 0.148 mol/L = 0.148 M (we approximate the volume of the solution to be equal to the volume of water, since solid Na2SO4 is assumed to occupy minimum volume).

Use the dilution equation to determine the concentrations of the next two solutions. The dilution equation is

M1*V1 = M2*V2

where M1 =0.148 M is the concentration of solution 1; V1 = 10 mL = volume of solution 1 taken and V2 = 50 mL = final volume of solution 2. Plug in values and obtain

(0.148 M)*(10 mL) = M2*(50 mL)

====> M2 = (0.148 M)*(10)/(50) = 0.0296 M.

Finally 15 mL of 0.0296 M solution 2 was taken and diluted to 25 mL. Use the dilution equation again to determine the concentration of solution 3 as

M1*V1 = M2*V2

====> (15 mL)*(0.0296 M) = M2*(25 mL)

====> M2 = (0.0296 M)*(15)/(25) = 0.01776 M.

The concentration of the most dilute Na2SO4 solution is 0.01776 M (ans).

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