You have 723 ml of 3.99 M HCl. Using a volumetric pipet, you take 326 ml of that solution and dilute it to 976 ml in a volumetric flask. Now you take 100.00 ml of that solution and dilute it to 119 ml in a volumetric flask. What is the concentration of hydrochloric acid in the final solution

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Answer:

1.12M

Explanation:

The initial concentration of the HCl solution is 3.99M. You are doing 2 dilutions:

In the first you take 326mL and dilute the solution to 976mL; that means you are diluting the solution:

976mL / 326mL = 2.99 times.

The concentration of this solution is:

3.99M / 2.99 = 1.33M

Now, in the second dilution, you take 100mL of the diluted solution and dilute it to 119mL. The concentration of this final solution is:

119mL / 100mL = 1.19 times.

1.33M / 1.19 =

1.12M

The final concentration of hydrochloric acid present is 1.11 M.

  • In the first dilution;

M1V1 = M2V2

M2 = 3.99 M × 326 ml/976 ml

M2 = 1.33 M

In the second dilution:

M1V1 = M2V2

M2 = 1.33 M × 100.00 ml/119 ml

M2 = 1.11 M

Hence, the final concentration of hydrochloric acid present is 1.11 M.

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