Prelab Assignment Experiment 8 Answers
1.
What is a colloid? Why is it difficult to filter a colloid?
A colloid is a dispersion of very tiny particles in a liquid. The particles are so small that they remain in
suspension, kept there by collisions with solvent molecules, rather than falling to the bottom of the
liquid as a precipitate. They are too small to be seen as individual particles with a conventional (light)
microscope, but they scatter light, thus making a liquid that contains them, such as milk, appear
cloudy. Colloids may be gaseous (for example smoke) or solid (for example, some types of
gemstones), but most commonly are found in the liquid phase.
Colloidal particles do not aggregate or flocculate (i.e., join together to form larger particles) because
they have a coating of charged ions, so as particles get closer they repel, and tend to then move
apart. This implies that if we add something to the solution that will neutralise the charge on the
surface of the colloids we should be able to get the colloidal particles to collide and thus precipitate.
This is correct – addition of a strong electrolyte often brings about flocculation.
Colloids are difficult to filter because the particles are so small that they pass through normal filter
paper or sintered glass funnels.
2.
Why is the solution of AgNO3 added slowly?
Slow addition tends to yield slightly larger colloidal particles which more readily flocculate, making
filtering more efficient. Also, slow addition reduces the local concentration of silver ions, which can
help to limit co-precipitation.
3.
Why must a large excess of AgNO3 be avoided?
To avoid co-precipitation.
4.
Why must the silver chloride be protected from light?
Many silver salts are fairly unstable, and can be decomposed by visible light. If the product starts to
decompose the yield of silver chloride will, of course, be reduced. The precipitate when first
generated should be a light creamy-yellow. This slowly darkens to a yellow-grey on exposure to light
and eventually to a dark grey or black if exposure is prolonged. The colour arises from the
generation of atomic silver, which is black.
5.
For simplicity assume that your unknown is NaCl. Calculate how many mL of the 5% AgNO3
solution (containing 5 g of silver nitrate per 100 mL of solution) must be added to 0.4020 g of the
NaCl solution sample to precipitate all the chloride.
The reaction between silver nitrate and sodium chloride is a simple one:
AgNO3 (aq) + NaCl (aq)  AgCl (s) + Na+ (aq) + NO3- (aq)
We see that one mole of NaCl requires one mole of AgNO3 for reaction, so if we work out
how many moles of chlorine there are in the sample we know how many moles of silver
nitrate are required for reaction.
The molecular weight of sodium chloride is 23 + 35.5 (AWs from the Periodic table)
= 58.5 g/mol. Thus, using moles=Wt/MW, the number of moles of NaCl in the sample is
0.4020/58.5 = 0.006872 moles.
Because of the 1:1 ratio of reactants as shown in the equation, this is also the number of
moles of silver nitrate in the required volume of solution.
The molecular weight of silver nitrate is 108 + 14 + 3 x 16 = 170 g/mol, so 5g (the amount in
100 mL) is 5/170 = 0.0294 moles.
Now it’s easiest to set up a pair of word equations:
100 mL solution contains 0.0294 moles silver nitrate
x mL solution contains 0.006872 moles silver nitrate
Solve however you like to find that x = 23.36 mL.
[Check: 23.36 mL contains {23.36/100} x 5g = 1.168g silver nitrate, which is 1.168/170 =
0.00687 moles as required.]