Could someone please explain how to plug these numbers into an equation? y (avg)
ID: 575794 • Letter: C
Question
Could someone please explain how to plug these numbers into an equation?
y (avg) = | 1.81 E+04 ASSIGNMENT After completing Part A in Experiment 4, a student obtained a trendline for the Beer's Law plot giving the following equation of absorbance, A, as a function of molar concen- tration, c, for the ion, FeSCN: A = 3752c-0.0141 The student then made up a series of five solutions of Fe3+ and SCN using 0.500 M HNO3(aq) as the solvent. Each solution was rnade using the volumes in liters of 2.00 × 10-3 M Fe(NO3)3 (to give FeM) and 2.00 × 10-3 M KSCN (to give SCN-) solutions shown in the table below. Volumes of 0.500 M HNO,(a) were added to bring the total volume of each solution to 20.00 mL. The number of moles of Fe+ and SCN in each solution can be found from their molarities and the volumes used to make up each sample solution. And since the total volume (0.0200 L) of each sample is also known, the initial molarities of Fe3+ and SCN- in each solution can be obtained by entering the correct formulas under the Sample solutions (Initial) columns. M(KSCN) M(Fe(NO3)3) V(Total), L0.02000 2.00E-032.00E-03 Sample solutions (Initial) V (KSCN), L V(Fe(NO3)3), L M (KSCN) M(Fe(NO3)3) 0.00200 0.004000.01000 0.00600 0.01000 0.00800 0.01000 0.01000 0.01000 0.01000 Given the initial (that is, before reaction to form FeSCN2+ occurs) concentrations of Fe3 and SCN- in each equilibrium concentra subtracting the equilibrium concentration of FeSCN2+ from the calculated initial con centrations. This works because for every one FeSCN2+ formed, one each of temd SCN is lost as shown in the equation for the reaction: h solution, the equilibrium concentrations can be found by tions can be fo EXCEL TUTORIAL B-5Explanation / Answer
Chemical equation,
Fe3+(aq) + SCN-(aq) <==> FeSCN^2+
From the trendline equation,
A = 3752c - 0.0141
molar absorptivity of FeSCN^2+ = 3752 M-1.cm-1
Beer's relation,
Absorbance = molar absorptivity (M-1.cm-1) x concentration (M) x path length (cm)
with path length = 1 cm
concentration of FeSCN^2+ = absorbance/molar absorptivity
Table : Initial concentration
Solution M(KSCN) M(Fe(NO3)3)
1 0.002 M x 0.002 L/0.020 L = 2 x 10^-4 M 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M
2 0.002 M x 0.004 L/0.020 L = 4 x 10^-4 M 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M
3 0.002 M x 0.006 L/0.020 L = 6 x 10^-4 M 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M
4 0.002 M x 0.008 L/0.020 L = 8 x 10^-4 M 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M
5 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M 0.002 M x 0.010 L/0.020 L = 1 x 10^-3 M
From absorbance equilibrium concentration of FeSCN^2+
Table
Solution Absorbance Equilibrium M(FeSCN^2+)
1 0.0995 0.0995/3752 = 2.65 x 10^-5 M
2 0.2060 0.2060/3752 = 5.5 x 10^-5 M
3 0.2872 0.2872/3752 = 7.65 x 10^-5 M
4 0.4306 0.4306/3752 = 1.15 x 10^-4 M
5 0.5575 0.5575/3752 = 1.5 x 10^-4 M
Equilibrium concentration of [Fe3+] and [SCN-] is initial concentration - equilibrium concentration [FeSCN^2+]
Table
Solution Equilibrium M(SCN-) Equilibrium M(Fe^3+)
1 2 x 10^-4 - 2.65 x 10^-5 = 1.73 x 10^-4 M 1 x 10^-3 - 2.65 x 10^-5 = 9.73 x 10^-4 M
2 4 x 10^-4 - 5.5 x 10^-5 = 3.45 x 10^-4 M 1 x 10^-3 - 5.5 x 10^-5 = 9.45 x 10^-4 M
3 6 x 10^-4 - 7.65 x 10^-5 = 5.23 x 10^-4 M 1 x 10^-3 - 7.65 x 10^-5 = 9.23 x 10^-4 M
4 8 x 10^-4 - 1.15 x 10^-5 = 6.85 x 10^-4 M 1 x 10^-3 - 1.15 x 10^-5 = 9.88 x 10^-4 M
5 1 x 10^-3 - 1.5 x 10^-4 = 8.5 x 10^-4 M 1 x 10^-3 - 1.5 x 10^-4 = 8.5 x 10^-4 M
From the above table,
Keq = [M(FeSCN^2+)]/[M(Fe3+)][M(SCN-)]
Solution Keq Equilibrium M(Fe^3+)
1 (2.65 x 10^-5)/(1.73 x 10^-4)(9.73 x 10^-4) = 157
2 (5.5 x 10^-5)/(3.45 x 10^-4)(9.45 x 10^-4) = 169
3 (7.65 x 10^-5)/(5.23 x 10^-4)(9.23 x 10^-4) = 158
4 (1.15 x 10^-5)/(7.88 x 10^-4)((9.88 x 10^-4) = 17
5 (1.5 x 10^-4)/(8.5 x 10^-4)(8.5 x 10^-4) = 207
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.