Hi I don’t even know how to start these 2 questions. I know you need the procedu
ID: 542222 • Letter: H
Question
Hi I don’t even know how to start these 2 questions. I know you need the procedure page for the solution mixes but I didn’t know if you needed the rest so I attached it anyways. Thank you!! Equilibriunm Introduction: Many chemical reactions can easily go forward or backward. The reaction shown below is one example. Fe+SCNFe(SCN)* In solution, Fe-reacts with SCN. to give the red ion Fe(SCN)" The Fe(SCN)2, can break apart to produce the original Fe ion and SCN:. In solution, both the forward and backward reactions are constantly occurring. The equilibrium constant K, of the reaction is given by the equation below [Fe+] [SCN] At equilibrium, the forward and backward reactions occur at the same rate, and the concentrations of the ions remain constant. This reaction is very fast, so the solution will reach equilibrium in less than one second. A general rule applies to situations where an equilibrium system is altered. Le Chatelier's principle states that when an equilibrium is disturbed, the reaction will shift in whichever direction will partially undo the disturbance. For instance, in the case above, adding Fe" to the solution caused the reaction to shift forward, which reduced the concentration of Fe". The principle also applies to heat. If the forward reaction is endothermic, adding heat to the reaction will made it shift forward, because the forward reaction will consume heat, so heat acts like a reactant. To find the value of Ke. you must determine the concentrations of all three ions at equilibrium. The best way to do this is to determine the initial concentrations of the ions, before any reaction occurred, and use this value to find the equilibrium concentrations of the three ions. You will be adding known amounts of Fe(NOs)s and KSCN to the solution, so it will be easy to calculate the initial concentrations of the Fe" and SCN ions. Remember that as you add these ions to the solution, you are diluting the solution. The equation for dilution is: after the reaction where M is the initial concentration of a solution, Vi is the initial volume, M2 is the final concen Fe3 is diluted to 20 mL, the new concentration of the solution will be 0.30 M. You will tration, and V2 is the final volume. For instance, if 10 mL a 0.60 M solution of not add any Fe(SCN) directly, so the initial concentration of this is zero. Once the Fe and SCN ions are added, the color of Fe(SCN) will appear. The concentration of Fe(SCNcan be determined by Beer's Law, which is as shown below 20Explanation / Answer
1) Use the dilution equation to find out the concentration of Fe3+ in the diluted solution. The dilution equation is
M1*V1 = M2*V2 where M1 = concentration of Fe3+ in the stock solution = 0.010 M; V1 = volume of stock solution taken; V2 = final volume of diluted solution = 6.00 mL and M2 = concentration of Fe3+ in the final diluted solution.
Take solution 1 as an example. We have V1 = 1.00 mL. Plug values in the dilution equation as below.
(1.0 mL)*(0.010 M) = (6.0 mL)*M2
====> M2 = 0.00167 M
The concentration of Fe3+ in solution#1 is 0.0167 M (ans).
Use the dilution equation to find out the concentrations of Fe3+ in solutions#2-5 and tabulate the results as below.
Solution #
Volume of 0.010 M Fe3+ taken (mL)
Concentration of Fe3+ in final solution (M)
1
1.00
0.00167
2
2.00
0.00333
3
0.50
0.00083
4
1.00
0.00167
5
1.00
0.00167
2) Use the dilution equation to find out the concentration of SCN- in the diluted solution. The dilution equation is
M1*V1 = M2*V2 where M1 = concentration of SCN- in the stock solution = 0.0020 M; V1 = volume of stock solution taken; V2 = final volume of diluted solution = 6.00 mL and M2 = concentration of SCN- in the final diluted solution.
Take solution 1 as an example. We have V1 = 1.00 mL. Plug values in the dilution equation as below.
(1.0 mL)*(0.0020 M) = (6.0 mL)*M2
====> M2 = 3.33*10-4 M
The concentration of SCN- in solution#1 is 3.33*10-4 M (ans).
Use the dilution equation to find out the concentrations of SCN- in solutions#2-5 and tabulate the results as below.
Solution #
Volume of 0.0020 M SCN- taken (mL)
Concentration of SCN- in final solution (M)
1
1.00
3.33*10-4
2
1.00
3.33*10-4
3
1.00
3.33*10-4
4
2.00
6.67*10-4
5
0.50
1.67*10-4
Solution #
Volume of 0.010 M Fe3+ taken (mL)
Concentration of Fe3+ in final solution (M)
1
1.00
0.00167
2
2.00
0.00333
3
0.50
0.00083
4
1.00
0.00167
5
1.00
0.00167