Exp 2: Volumetric Analysis

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

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Experiment 2 Virtual Lab Tutorial: Acid Base Titration

Step 2: Calculations for the standardization of NaOH

In this step, we will use the 0.09996 M KHP standard solution prepared in step 1, to determine the concentration of the NaOH solution with 4 significant figures precision. The concentration of the NaOH available in the laboratory is approximately 1.0 M, which is 10 times the concentration of the KHP. In a titration, the best results are obtained when similar volumes of the titrant and titrated sample are used. A 25.00 mL sample of the 0.09996 M standard solution is going to be used in the titration, so the ideal volume of the titrant (NaOH) used will be 25 mL.

Tutor 3: Calculations for diluting NaOH

If we were to use 1.0 M NaOH solution in this titration, what volume would we use to complete the titration?

Hint

2.5 mL of 1.0 M NaOH

25 mL of 1.0 M NaOH

250 mL of 1.0 M NaOH

Good Job!

That's not quite right.

Hint:

KHP reacts with NaOH in 1:1 ratio. Since NaOH is 10x more concentrated than KHP, you will need 10x less of it to reach the completion of the reaction.

Therefore, to obtain best results in our titration, we must dilute the NaOH solution 10 times before we titrate it with the KHP standard solution, so that similar amounts of NaOH and KHP will be used.

What volume (in mL) of 1.0M NaOH and deionized water are required to prepare 500 mL of 0.1 M NaOH solution?

Hint

mL 1.0 M NaOH

mL deionized water

Good Job!

That's not quite right.

Hint:

The volume of NaOH used to perform the dilution will contain a certain number of moles of NaOH. Keeping in mind that moles NaOH = MNaOH x VNaOH, the number of moles of NaOH before the dilution, M1 x V1, is the same as the number of moles of NaOH after the dilution, M2 x V2. Therefore, M1 x V1 = M2 x V2.

get next hint

Hint:

Rearranging the above equation to solve for the volume

V1 = (M2 x V2) / M1.

get previous hint

get next hint

Hint:

Substituting the known values to the above equation, we obtain

V1 = (0.1 M x 0.5 L) / 1.0M = 0.05 L or 50 mL.

get previous hint

Hint:

Since the total volume required is 500 mL and we are using 50 mL of the 1.0 M NaOH, we require 450 mL of water.

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. We only need to report the concentration of the NaOH solution after it has been standardized. Therefore, we can dilute the 1.0 M stock NaOH solution without observing the precision rules required in a quantitative dilution.

Hint:

We only need to report the concentration of the NaOH solution after it has been standardized. Therefore, we can dilute the 1.0 M stock NaOH solution without observing the precision rules required in a quantitative dilution.

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