In this lab you will need to prepare solutions using dilutions. Starting with the stock 0.300 M NaOH solution, how would you prepare a 0.050 M NaOH solution (using 0.300 M NaCl as the diluent)? To prepare 24 mL of 0.050 M NaOH solution, you would add mL of 0.300 M NaOH stock solution and mL of 0.300 M NaCl solution.

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05-02-2020
Chemistry

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In this lab you will need to prepare solutions using dilutions. Starting with the stock 0.300 M NaOH solution, how would you prepare a 0.050 M NaOH solution (using 0.300 M NaCl as the diluent)? To prepare 24 mL of 0.050 M NaOH solution, you would add mL of 0.300 M NaOH stock solution and mL of 0.300 M NaCl solution.

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Tringa0 Tringa0 · Answer 1 · 2020-02-06T10:13:17+01:00

Answer:

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

Explanation:

Molarity of the NaOH solution = [tex]M_1=0.300 M[/tex]

Volume of the NaOH solution = [tex]V_1=?[/tex]

Molarity of the NaOH solution after dilution= [tex]M_2=0.050 M[/tex]

Volume of the NaOH solution after dilution= [tex]V_2=24 mmL[/tex]

[tex]M_1V_1=M_2V_2[/tex] (Dilution )

[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.050 M\times 24 mL}{0.300 M}=4 mL[/tex]

Volume of NaCl solution of 0.300 M = 24 mL - 4 mL = 20 mL

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

Eduard22sly Eduard22sly · Answer 2 · 2022-01-25T11:50:51+01:00

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution

Determination of the volume of the stock solution of NaOH needed

•Molarity of stock solution (M₁) = 0.3 M

•Molarity of diluted solution (M₂) = 0.05 M

•Volume of diluted solution (V₂) = 24 mL

•Volume of stock solution needed (V₁) =?

Using the dilution formula, the volume of the stock solution needed can be obtained as follow:

M₁V₁ = M₂V₂

0.3 × V₁ = 0.05 × 24

0.3 × V₁ = 1.2

Divide both side by 0.3

V₁ = 1.2 / 0.3

V₁ = 4 mL

Determination of the volume of NaCl needed

•Volume of NaOH needed = 4 mL

•Volume of diluted solution of NaOH = 24 mL

•Volume of NaCl needed =?

Volume of NaCl needed = (Volume of diluted solution of Na) – (Volume of NaOH needed)

Volume of NaCl needed = 24 – 4

Volume of NaCl needed = 20 mL

Therefore, we can conclude as follow:

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

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