Calculate [SCN-]_initial for the standard solutions given in Table 1. As explain
ID: 490331 • Letter: C
Question
Calculate [SCN-]_initial for the standard solutions given in Table 1. As explained in the Background section, the equilibrium for the reaction Fe^3+ rightarrow FeSCN^2+ was intentionally shifted far to the right in preparing the Standard solutions so [SCN^-]_initial for each of the standard solutions = [FeSCN^2+]_eq. Copy the values into notebook for part IV of the lab. Calculate [SCN]_initial and [Fe^3+]_initial for the experimental solutions given in Table 2. Copy the values into your notebook for part V of the lab. Using the spectroscopic method described in this lab, a student determined that [FeSCN^2+]_eq for one sample was 1.4 times 10^-4 M. For this same sample, [Fe^3+]_initial was 1.0 times 10^-3 M and [SCN^-]_initial was also 1.0 times 10^-3 M. Use this information to find [Fe^3+]_eq and [SCN^-]_eq. From the values in problem 3 of [FeSCN^2+]_eq, and [SCN^-]_eq, find K_eq for formation of FeSCN^2+ from Fe^3+ and SCN^-.Explanation / Answer
1) Let’s work with solution S2 as an example. We have 1.00 mL of 2.00*10-3 M NaSCN and the total volume of the solution was made up to 50.00 mL.
Use the dilution equation to determine [SCN-]initial as below:
M1*V1 = M2*V2 where M1 = 2.00*10-3 M; V1= 1.00 mL, V2 = 50.00 mL. Plug in values and obtain [SCN-]initial
(1.00 mL)*(2.00*10-3 M) = (50.00 mL)*[SCN-]initial
===> [SCN-]initial = (1.00*2.00*10-3 M)/(50.00) = 4.0*10-5 M (ans).
Fill up the table as given.
Solutions
2.00*10-3 M NaSCN in 0.10 M HNO3 (mL)
Total volume of solution (mL)
[SCN-]initial (M)
S1 Blank
0
50.00
0
S2
1.00
50.00
4.0*10-5
S3
2.00
50.00
8.0*10-5
S4
3.00
50.00
1.2*10-4
S5
4.00
50.00
1.6*10-4
S60
5.00
50.00
2.0*10-4
2) I need Table 2 to answer the question.
3) Given [FeSCN2+]eq = 1.4*10-4 M; [Fe3+]initial = 1.0*10-3 M and [SCN-]initial = 1.0*10-3 M; therefore,
[Fe3+]eq = [Fe3+]initial – [FeSCN2+]eq = (1.0*10-3 M) – (1.4*10-4 M) = 8.6*10-4 M (ans).
[SCN-]eq = [SCN-]initial – [FeSCN2+]eq = (1.0*10-3 M) – (1.4*10-4 M) = 8.6*10-4 M (ans).
4) The equilibrium reaction is
The equilibrium constant is given by
Therefore, Keq = (1.4*10-4)/(8.6*10-4).(8.6*10-4) = (1.4*10-4)/(8.6*10-4)2 = 189.291 189.3 (ans).
Solutions
2.00*10-3 M NaSCN in 0.10 M HNO3 (mL)
Total volume of solution (mL)
[SCN-]initial (M)
S1 Blank
0
50.00
0
S2
1.00
50.00
4.0*10-5
S3
2.00
50.00
8.0*10-5
S4
3.00
50.00
1.2*10-4
S5
4.00
50.00
1.6*10-4
S60
5.00
50.00
2.0*10-4
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