Precipitate Reactions
Precipitate reactions
In reactions where a precipitate is formed, the amount of precipitate formed in a period of time can be used as a measure of the reaction rate. For example, when sodium thiosulfate reacts with an acid, a yellow precipitate of sulfur is formed. The reaction is as follows:
Fact:
When the reaction between two liquids forms an insoluble compound, solid material will form in the liquid. This solid is called a precipitate.
\(\text{Na}_{2}\text{S}_{2}\text{O}_{3}(\text{aq}) + 2\text{HCl}(\text{aq})\) \(\to\) \(2\text{NaCl}(\text{aq}) + \text{SO}_{2}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) + \text{S}(\text{s})\)
One way to estimate the average rate of this reaction is to carry out the investigation in a conical flask and to place a piece of paper with a black cross underneath the bottom of the flask. At the beginning of the reaction, the cross will be clearly visible when you look into the flask (see figure below). However, as the reaction progresses and more precipitate is formed, the cross will gradually become less clear and will eventually disappear altogether. Measuring the time that it takes for this to happen will give an idea of the reaction rate. Note that it is not possible to collect the \(\text{SO}_{2}\) gas that is produced in the reaction, because it is very soluble in water.
At the beginning of the reaction between sodium thiosulfate and hydrochloric acid, when no precipitate has been formed, the cross at the bottom of the conical flask can be clearly seen. As the precipitate forms less of it can be seen.
For the reaction of sodium thiosulfate with hydrochloric acid it would be good to break your class up into groups and assign different experiments to them at this point. They can report back to the rest of the class.
This experiment is recommended for informal assessment in CAPS. Remember to remind the learners about laboratory safety procedures, especially when handling acids.
Optional Experiment: Measuring reaction rates
Aim
To measure the effect of concentration on the average rate of a reaction.
Apparatus
-
\(\text{300}\) \(\text{cm$^{3}$}\) of sodium thiosulfate \(\text{Na}_{2}\text{S}_{2}\text{O}_{3}\) solution. (Prepare a solution of sodium thiosulfate by adding \(\text{12}\) \(\text{g}\) of \(\text{Na}_{2}\text{S}_{2}\text{O}_{3}\) to \(\text{300}\) \(\text{cm$^{3}$}\) of water). This is solution 'A'.
-
\(\text{300}\) \(\text{cm$^{3}$}\) of water
-
\(\text{100}\) \(\text{cm$^{3}$}\) of \(\text{1}\):\(\text{10}\) dilute hydrochloric acid. This is solution 'B'.
-
Six \(\text{100}\) \(\text{cm$^{3}$}\) glass beakers, measuring cylinders, paper and marking pen, stopwatch or timer
Method
Warning:
Do not get hydrochloric acid (\(\text{HCl}\)) on your hands. We suggest you use gloves and safety glasses whenever handling acids and that you handle with care.
One way to measure the average rate of this reaction is to place a piece of paper with a cross underneath the reaction beaker to see how long it takes until the cross cannot be seen due to the formation of the sulfur precipitate.
-
Set up six beakers on a flat surface and label them 1 to 6.
-
Pour \(\text{60}\) \(\text{cm$^{3}$}\) solution A into the first beaker and add \(\text{20}\) \(\text{cm$^{3}$}\) of water
-
Place the beaker on a piece of paper with a large black cross on it.
-
Use the measuring cylinder to measure \(\text{10}\) \(\text{cm$^{3}$}\) \(\text{HCl}\). Now add this \(\text{HCl}\) to the solution that is already in the first beaker (NB: Make sure that you always clean the measuring cylinder you have used before using it for another chemical).
-
Using a stopwatch with seconds, write down the time it takes for the precipitate that forms to block out the cross.
-
Now measure \(\text{50}\) \(\text{cm$^{3}$}\) of solution A into the second beaker and add \(\text{30}\) \(\text{cm$^{3}$}\) of water. Place the beaker over the black cross on the paper. To this second beaker, add \(\text{10}\) \(\text{cm$^{3}$}\) \(\text{HCl}\), time the reaction and write down the results as you did before.
-
Continue the experiment by diluting solution A as shown below.
| Beaker | Solution A (\(\text{cm$^{3}$}\)) | Water (\(\text{cm$^{3}$}\)) | Solution B (\(\text{cm$^{3}$}\)) | Time (s) |
| \(\text{1}\) | \(\text{60}\) | \(\text{20}\) | \(\text{10}\) | |
| \(\text{2}\) | \(\text{50}\) | \(\text{30}\) | \(\text{10}\) | |
| \(\text{3}\) | \(\text{40}\) | \(\text{40}\) | \(\text{10}\) | |
| \(\text{4}\) | \(\text{30}\) | \(\text{50}\) | \(\text{10}\) | |
| \(\text{5}\) | \(\text{20}\) | \(\text{60}\) | \(\text{10}\) | |
| \(\text{6}\) | \(\text{10}\) | \(\text{70}\) | \(\text{10}\) |
The equation for the reaction between sodium thiosulfate and hydrochloric acid is:
\(\text{Na}_{2}\text{S}_{2}\text{O}_{3}(\text{aq}) + 2\text{HCl}(\text{aq})\) \(\to\) \(2\text{NaCl}(\text{aq}) + \text{SO}_{2}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) + \text{S}(\text{s})\)
Results
-
Calculate the reaction rate in each beaker. Remember that:
rate of the formation of product = \(\dfrac{\text{moles product formed}}{\text{reaction time (s)}}\)
In this experiment you are stopping each experiment when the same approximate amount of precipitate is formed (the cross is blocked out by precipitate). So a relative reaction rate can be determined using the following equation:
reaction rate = \(\dfrac{1}{\text{time (s)}}\)
-
Represent your results on a graph. Concentration will be on the x-axis and reaction rate on the y-axis. Note that the original volume of \(\text{Na}_{2}\text{S}_{2}\text{O}_{3}\) can be used as a measure of concentration.
-
Why was it important to keep the volume of \(\text{HCl}\) constant?
-
Describe the relationship between concentration and reaction rate.
Conclusions
The higher the concentration of the reactants, the faster the average reaction rate.
This lesson is part of:
Energy and Chemical Reactions