Exp 2: Volumetric Analysis

Experiment 2 Virtual Lab Tutorial >> Acid Base Titration >> Step 2: Tutor 4

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Experiment 2 Virtual Lab Tutorial: Acid Base Titration

Step 2: Calculations for the standardization of NaOH

Now that we have prepared a 0.1 M solution of NaOH, we can find its exact concentration, with 4 significant figures precision, by performing a titration with the KHP standard solution.

Tutor 4: Determining the concentration of NaOH

In a titration of 25.00 mL of 0.09996 M KHP standard solution with 0.1 M NaOH, 24.29 mL of NaOH was used to reach the endpoint. What is the concentration of NaOH to 4 significant figures?

Hint

M NaOH

Please give your answer to 4 significant figures.

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That's not quite right.

Hint:

Since KHP reacts with NaOH in 1:1 ratio, then the number of moles of KHP will be 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

MKHP x VKHP = MNaOH x VNaOH

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Hint:

Rearranging the above equation, to solve for M of NaOH, we obtain:

MNaOH = MKHP x VKHP / VNaOH

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Hint:

The result is MNaOH = (0.09996 M x 25.00 mL) / 24.29 mL = 0.1029 M

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