Exp 2: Volumetric Analysis

Experiment 2 Virtual Lab Tutorial >> Acid Base Titration >> Step 3: Tutor 5

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Experiment 2 Virtual Lab Tutorial: Acid Base Titration

Step 3: Calculations for the standardization of unknown HCl

In this step, we will use the 0.1029 M NaOH solution to determine the concentration of the unknown HCl solution with 4 significant figures precision. The concentration of the unknown HCl is approximately 1.0 M, which is 10 times the concentration of the NaOH. Remember, in a titration, the best results are obtained when similar volumes of the titrant and titrated sample are used, so this time we will need to dilute the HCl solution 10 times.

Tutor 5: Determining the concentration of the diluted unknown HCl

What is the appropriate glassware for performing this dilution in the laboratory?

Hint

A volumetric pipet and a volumetric flask

A 50 mL graduated cylinder and a bottle to hold the diluted solution

Good Job!

That's not quite right. Our goal in this experiment is to determine the concentration of the 1.0 M HCl solution and not just the diluted solution (as was the case with the NaOH). Therefore we must dilute the 1.0 M HCl quantitatively to maintain the required precision.

Hint:

Our goal in this experiment is to determine the concentration of the 1.0 M HCl solution and not just the diluted solution (as was the case with the NaOH). Therefore we must dilute the 1.0 M HCl quantitatively to maintain the required precision.

In a titration using a 25.00 mL sample of the 0.1 M diluted HCl solution with 0.1029 M NaOH, it took 22.31 mL of NaOH to reach the endpoint. What is the concentration of the diluted HCl solution to 4 significant figures?

Hint

M HCl

Please give your answer to 4 significant figures.

Good Job!

That's not quite right.

Hint:

Since HCl reacts with NaOH in 1:1 ratio, then the number of moles of HCl will equal to the number of moles of NaOH when the reaction reaches completion (the end point of the titration).

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Hint:

Keeping in mind that moles = Molarity x Volume, and

MHCl x VHCl = MNaOH x VNaOH

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Hint:

Rearranging the above equation, we obtain:

MHCl = MNaOH x VNaOH / VHCl

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Hint:

The result is MHCl = (0.1029 M x 22.31 mL) / 25.00 mL = 0.09183 M

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