Question

2. (Solution Preparation 8 points) You have a standardized aqueous solution with a Cu concentration of 23.1 ppm (in terms of mass) and a density = 1.03 g/ml at 298 K. From this standardized solution you need to create the dilutions specified below. Make sure to report all quantities with the proper number of significant digits and appropriate units, and show the process you employed to calculate these dilutions to receive credit. You do not need to propagate uncertainty in this problem. a. (4 points) 150.0 mL of 10.0 ppm Cu solution. b. (4 points) 100.0 mL of 100. nM solution

I really can't seem to get a hang of the PPM calculations. Pretend like you're explaining this to a 5 year old if you can.

Answer 1

When the solvent is added in a solution, than the volume of solution increases and the concentration decreases. This is process of dilution. The concentration and the volume of dilute and concentrated solution are given as follows: C_1 V_1 Concentrated solution = C_2 V_2 Dilute solution Given that Standardized solution with Cu2 + = 23.1 ppm by mass Density = 1.03 g/ ml a. 150.0 ml of 10.0 ppm Cu 2 + solution Here C1 = 23.1 ppm V! = ? in ml C2 = 10.2 ppm and V2 = 150.0 ml Then: C1 V1 = C2 V2 V1 = 10.0 ppm * 150.0 ml /23.1 ppm = 64. 94 ml = 65 ml Now take 65 ml of 23.1 ppm solution and dilute it up to 150 ml b. first calculate the morality of 23.1 ppm Cu 2 + solution as follows: ppm = 1 gm solute per 1 kg solution here density = 1.03 g/ ml density = mass/ volume for volume = mass/ density for 1.0 kg or 1000 g:

first calculate the molarity of 23.1 ppm Cu2+ solution as follows:

ppm = 1 mg solute per 1 kg solution

here density = 1.03 g/ ml

density = mass/ volume

for volume = mass / density

for 1.0 kg or 1000 g:

volume = 1000 g/ 1.03 g/ ml

= 970.8 ml

= 0.9708 L

Thus

ppm = 23.1 mg solute per 0.9708 liter of solution

or 0.0231 g per 0.9708 liter of solution

now Divide by the atomic weight for Cu ion:

0.0231 g/ 0.9708 L divided by 63.55 g/mol =0. 374 mM

Then

100.0 ml of 100 mM Cu2 + solution

Here C1 =0.374 mM

V1= ? in ml

C2 = 100 nM or 0.0001mM   and V2= 100.0 ml

Then;

C1V1= C2V2

V1= 0.0001 m M * 100.0 ml / 0.374 mM

=0.0267 ml

= 0.03 ml

Now take 0.03 ml of 23.1 ppm solution and dilute it up to 100.0ml