Calculate the volume of 1.0 x 10-4 M CV+ solution that needs to be added to a 25.0 mL volumetric flask and diluted with deionized (DI) water in order to prepare a calibration standard solution with a concentration of 1.0 x 10-5 M CV+. As part of your preparation for performing this experiment, repeat this calculation for each of the calibration standards you will need to prepare in Part I A of the experiment and record the information in your lab notebook so that you have it ready during the lab session. Also calculate the volume of 1.0 x 10-4 M CV+ solution required to prepare 10 mL of 3.0 x 10-5 M CV+ solution to be used in Part II B of the experiment and record the information in your lab notebook.

Where Vc and Cc are the volume and concentration respectively of concentrated solution (higher concentration) whereas Vd and Cd are volume and concentration of diluted solution (lower concentration).

In both problems we want to prepare a diluted solution and we know the final concentration (Cd) and final volume (Vd) and the initial concentration (Cc).

In first part, we have: Cc= 1.0 10⁻⁴ M; Vd= 25 ml; Cd= 1.0 10⁻⁵ M

Vc= Vd x Cd / Cc= (25 ml x 1.0 10⁻⁵ M)/1.0 10⁻⁴ M = 2.5 ml

Notice that Cc/Cd= 1.0 10⁻⁴ M/Cd= 1.0 10⁻⁵ M= 10 (so, we have to dilute the solution 10 times, and for this we have to take a volume 10 times lower than the final volume).

To prepare the solution, we take 2.5 ml of 1.0 10⁻⁵ M CV+, we dispense the volume in a 25 ml volumetric flask and then we add water until complete 25 ml (aproximately 22.5 ml of water).

In the second part is the same. We have: Vd= 10 ml; Cc= 1.0 10⁻⁴M; Cd= 3.0 10⁻⁵M.

Vc= Vd x Cd / Cc= (10 ml x 3.0 10⁻⁵ M)/1.0 10⁻⁴ M = 3 ml

To prepare the solution, we take 3 ml of 3.0 10⁻⁵ M CV+, we dispense the volume in a 10 ml volumetric flask and then we add water until we complete 10 ml (aproximately 7 ml of water).

tutorareej tutorareej · Answer 2 · 2021-10-09T11:05:19+02:00

Answer:

The volume of solution required to make final concentration of [tex]\rm 1\;\times\;10^-^5[/tex] is 2.5 ml and for [tex]\rm 3\;\times\;10^-^5[/tex] is 3 ml.

Explanation:

Calibration standard solution is the solution prepared of the known concentration for the plotting of standard curve.

For preparing the solution of known concentration and volume, the equation imputed is :

[tex]\rm M_1V_1\;=\;M_2V_2[/tex]

where [tex]\rm M_1[/tex] and [tex]\rm V_1[/tex] are molarity and volume of known solution

[tex]\rm M_2[/tex] and [tex]\rm V_2[/tex] are molarity and volume of solution to be prepared.

Given, volume to be prepared [tex]\rm V_2[/tex] : 25 ml

concentration to be prepared [tex]\rm M_2[/tex] : [tex]\rm 1\;\times\;10^-^5[/tex] M

concentration of given solution [tex]\rm M_1[/tex] : [tex]\rm 1\;\times\;10^-^4[/tex] M

[tex]\rm 1\;\times\;10^-^4[/tex] [tex]\times[/tex] [tex]\rm V_1[/tex] = [tex]\rm 1\;\times\;10^-^5[/tex] [tex]\times[/tex] 25

Volume of solution required is 2.5 ml.

The volume required to prepare 10 ml solution of [tex]\rm 3\;\times\;10^-^5[/tex] M is:

[tex]\rm 1\;\times\;10^-^4\;\times\;V_1[/tex] = [tex]\rm 3\;\times\;10^-5\;\times\;10[/tex]

Volume = 3 ml.

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