Give learners a solution in water of a solid organic compound known to be a weak, monoprotic (monobasic) acid

They can find out and list as much information as possible about a weak acid from experimental results shown.

## Introduction

Teachers who have not used the problems before should read the section Using the problems before starting.

## Prior knowledge

Solution chemistry and in particular weak acids (K_{a}, pK_{a} and titration curves). A detailed knowledge is unnecessary as students are encouraged to consult textbooks and data books during the exercise.

## Resources

Graph paper; data books and textbooks for reference; and scientific calculators should be available at the start of the exercise.

## Method

The equivalence point of the titration can be determined by plotting pH against volume of alkali added. This occurred at 15.00 cm^{3} of exactly 0.1 mol dm^{–}^{3} NaOH.

The following information can now be determined:

- Total acid (HA) concentration = 0.100 x 15.00/25.0 = 0.0600 mol dm
^{–}^{3} _{a}= pH at half neutralisation = 4.20 (from graph)

therefore K_{a} = 6.31 x 10^{–}^{5}

- pK
^{b}of conjugate base (A) = 14^{–}^{4}.20

= 9.80

therefore K_{b} = 1.58 x 10^{–}^{10}

- The pH of the original solution (before any alkali is added) is found as follows:

K_{a} = [H^{+}]^{2}/[HA]

[H^{+}]^{2} = K_{a} x [HA]

= 6.31 x 10^{–}^{5} x 0.0600 (assuming that [HA] equals initial concentration of the acid)

[H^{+}] = 1.95 x 10^{–}^{3}

pH = 2.71

- The degree of dissociation of the acid (á) before any alkali is added is given by:

α = [H+]/0.0600

= 3.25 x 10^{–}^{2}

- The molar mass of the acid can be calculated as follows:

The original acid solution is 0.0600 mol dm^{–}^{3} (from **1 ** above). Exactly 100 cm^{3}

of it contains 0.00600 moles and this has a mass of 0.72 g.

Thus one mole of the acid has a mass of 0.72/0.00600 = 120 g.

**2**and**6**should allow the students to identify the acid as

benzoic acid (pK_{a} = 4.19, molecular mass = 122.13).

These are only examples of what students can calculate from the information given.

## Suggested approach

During trialling the following instructions were given to students and proved to be extremely effective:

- Working as a group discuss the task and carry it out. You can if you wish divide the work amongst you and then gather the results together. Such discussion can play a vital part in working out a solution to an open ended problem where divergent thinking is required. Several minds working on a problem together can stimulate ideas that one on its own could not manage. A few minutes should be spent discussing the task at the start of the exercise, with further discussion as required.
- Working as a group, prepare a short (ca 5-minute maximum) presentation to give to the rest of the class. If possible all group members should take part: any method of presentation (such as a blackboard, overhead projector, etc) can be used.

Outline the problem, describe what you did and explain your calculations and conclusions but do not give too much detail. After the presentation, be prepared to accept and answer questions and to discuss what you did with the rest of the class.

## Background information

To calculate the pH during titration of a weak acid with sodium hydroxide solution.

K_{a} = [H^{+}][A]/[HA]

Where [H^{+}], [A] and [HA] are all equilibrium concentrations.

[H^{+}]= Ka x [HA]/[A] or pH = pK^{a} + log([A]/[HA])

### Downloads

#### On the acid trail - creative problem solving

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### Additional information

This resource is part of our Creative problem-solving in chemistry collection.

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