Calculate the volume (in mL) of 0.00500 M KCl solution that needs to be added to a 50.0 mL volumetric flask and diluted with deionized (DI) water in order to prepare a calibration standard solution with a concentration of 1.50 x 10-4 M KCl. As part of your preparation for performing this experiment, repeat this calculation for each of the calibration standards you will need to prepare and record the information in your lab notebook so that you have it ready during the lab session.

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Explanation:

As the given data is as follows.

     [tex]M_{1}[/tex] = 0.005 M,       [tex]M_{2} = 1.50 \times 10^{-4}[/tex]

     [tex]V_{1}[/tex] = ? mL,        [tex]V_{2}[/tex] = 50.0 mL

When the number of moles of the species are equal then the formula will be as follows.

                     [tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

          [tex]0.005 \times V_{2} = 1.50 \times 10^{-4} \times 50.0 mL[/tex]

                  [tex]V_{2}[/tex] = 1.50 mL

Thus, we can conclude that volume that needs to be added is 1.50 mL.

                                 

The molarity of the substance is the ratio of the moles and the volume. The volume in mL needed of potassium chloride in the volumetric flask is 1.50 mL.

What is the relation between the molarity dilution?

Molarity and dilution are inversely proportional to each other as increased dilution decreases the molarity of the substance.

The molarity and the volume can be shown as,

[tex]\rm M_{1}V_{1} = M_{2}V_{2}[/tex]

Where,

[tex]\rm M_{1}[/tex] = 0.005 M

[tex]\rm M_{2}[/tex] = [tex]1.50 \times 10^{-4}[/tex]

[tex]\rm V_{1}[/tex]  = ?

[tex]\rm V_{2}[/tex] = 50.0 mL

Substituting values in the above equation:

[tex]\begin{aligned} \rm V_{2} &= \dfrac{\rm M_{1}V_{1}}{\rm M_{2}}\\\\&= \dfrac{0.005 \times 1.50 \times 10^{-4}}{50}\\\\&= 1.50\;\rm mL\end{aligned}[/tex]

Therefore, 1.50 mL of 0.00500 M KCl solution is needed to be added to the 50 mL of the volumetric flask.

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