In preparation for thinking about this system during lab, write an explanation i
ID: 1050252 • Letter: I
Question
In preparation for thinking about this system during lab, write an explanation in your lab notebook of your own understanding of what you think will happen in three experiments (labeled A, 0, and C in the table below) when different amounts of HAuC_4L(aq) and Na_3C_6H_5O_7(aq) are mixed at an elevated temperature. Describe what you expect to see occur at the macroscopic level as you produce Au(s) according to the balanced chemical reaction above. In other words, explain the observable changes you think you will see as HAuCl_4(aq) is converted into Au(s). Describe what differences you would expect to see (if any) among mixtures A, B, and C. Explain your reasoning for the expected differences (or lack of difference).Explanation / Answer
1. 4 moles of HAuCl4 reacts with 3 moles of Na3C6H5O7 which in turn gives 4 moles of Au
In bulk at the macroscale, the element of gold is gold colored, but at the in particles nanoscale, the element of gold is red to purple in color.
The experiment is related to the formation of nano particles of gold.
On mixing a gold solution with a sodium citrate solution we will see a color change that lets us know the gold particles are forming. The color of the particle solution depends on the size and shape of the particles being formed. The mixture will change from colorless to blue to red as the nanoparticles are produced and aggregated i.e., joined together.
Before the addition of the sodium citrate (reducing agent), the gold in solution is in the Au+3 form. When the reducing agent is added, gold atoms are formed in the solution, and their concentration rises rapidly until the solution exceeds saturation.
The ionic NaCl solution should cause the gold particles to aggregate and the solution will change from red to blue.The solution formed is a colliodal suspension of gold particles.
2. 25 X 10-6 mol of HAuCl4 needs (3/4) X 25 X 10-6 moles of Na3C6H5O7 = 18.75 X 10-6 moles of Na3C6H5O7 to react completely and gives 25 X 10-6 moles of Au.
As we see the mixtures A,B and C , only A and B ,HAuCl4 reacts completely and in both the cases Na3C6H5O7 is present in excess and in case C, HAuCl4 does not react completely because the concentration of Na3C6H5O7 is less than 18.75 X 10-6 moles(i.e., 13 X 10-6 moles). In case C, HAuCl4 is the excess reagent.The amount of HAuCl4 reacted is (4/3) X 13 X 10-6 = 17.33 X 10-6 moles and the amount of Au formed is 17.33 X 10-6 moles.
The gold particles formed in case A are seen clearly and in case B and C due to the presence of the excess reagents, the particles are covered by them.