TUTORIAL – Lab: Experiment 9 Practice Questions

Question 1: Explain why the beaker used to obtain the stock solution has to be clean and dry.

Answer to Question 1:

Here is a wet beaker. What would the water that is adhering on the side of the beaker do to the stock solution?

When the stock solution is poured into the wet beaker, the water will cause an INCREASE in the volume of the stock solution that is poured in. As a result of the transfer of the stock solution into the beaker the concentration of the stock solution will be DILUTED by some unknown quantity.

NOTE: If the objective of the analysis is to determine the concentration of the stock solution, this will result in error because the analysis will be carried out on the DILUTED stock solution (as a result of the wet beaker) instead of the concentration of the stock solution that is in the bottle!

The beaker that is receiving the stock solution MUST BE CLEAN AND DRY so that the concentration of the stock solution would not be altered as a result of transferring of the solution to the beaker.

Question 2: A titration is carried out to determine the concentration of nitric acid in an old bottle as the label has become unreadable.
(a) Write the balanced reaction between nitric acid and potassium hydroxide.

Answer to Question 2:

This is an acid-base reaction. The method to determine the concentration of the nitric acid is known as an acid-base titration. It involves using a base, potassium hydroxide, KOH, to react with the nitric acid, HNO3. When the acid is neutralized by the base, the concentration of the acid can be calculated (see Question 2(b)).

The reaction is

KOH (aq) + HNO3 (aq) KNO3 (aq) + H2O (l)

Question 2: A titration is carried out to determine the concentration of nitric acid in an old bottle as the label has become unreadable.
(b) What is the nitric acid concentration if 58.40 mL of 0.2500 M potassium hydroxide is required to titrate a 20.00 mL sample of the acid?
To carry out the titration, we would:

  1. Fill the burette with potassium hydroxide, KOH, and
  2. Pipette 20.00 mL of nitric acid HNO3 in the Erlenmeyer flask.
  3. Add a few drops of phenolpthalein, a colour indicator, to the Erlenmeyer flask to indicate when the acid in the Erlenmeyer flask is neutralized (i.e. no more acid is left in the Erlenmeyer flask).

    Phenolpthalein is colourless in acidic solution.

    Phenolpthalein is pink in basic solution.

  4. Titration endpoint is reached when the solution in the Erlenmeyer flask is pale pink.

Read the question carefully. It tells us information about the base and the acid. It is really important not to mix up these information!

Information about the base, potassium hydroxide, KOH:

  • Volume = 58.40 mL
  • Concentration = 0.2500 M

Information about the nitric acid, HNO3:

  • Volume = 20.00 mL
  • Concentration = ? M (We want to calculate this!)

We know that when 58.40 mL of the 0.2500 M potassium hydroxide, KOH, is added to the Erlenmeyer flask, the acid in the Erlenmeyer flask is neutralized. This is known as the endpoint of the titration. We know that the endpoint is reached because the solution in the Erlenmeyer flask becomes pale pink in colour.

Let’s use the expanded roadmap to calculate the concentration of the HNO3 from the amount (i.e. number of moles) of potassium hydroxide, KOH, that is added to the Erlenmeyer flask.


(0.2500 moles/L KOH)(0.05840 L KOH) = 0.01460 mole of KOH

This tells us that at the endpoint of the titration, 0.01460 mole of KOH has been added to the Erlenmeyer flask.

From the balanced equation in part (a),
1 KOH (aq) + 1 HNO3 (aq) KNO3 (aq) + H2O (l)

The mole ratio of KOH to HNO3 is 1:1.

Therefore, since 0.01460 mole of KOH is used in the titration to bring the reaction to the the endpoint, there must have been 0.01460 mole of HNO3 in the 20.00 mL of HNO3 in the Erlenmeyer flask.

Using the expanded road map again, we can calculate the concentration of HNO3.

Since the volume of the acid solution used is 20.00 mL,

Concentration of the HNO3 = (0.01460 mole HNO3 / 0.02000 L HNO3) = 0.7300 M HNO3

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