LAB INFORMATION:
Initial
Fe3+ =2.00 x 10-3 M
SCN- = 2.00 x 10-3 M
FeSCN3+ = 1.50 x 10-4 M
| MIXTURE | ABSORBANCE | FeSCN2+ |
|---|---|---|
| 1 |
0.149 |
3.08 x 10^-5 |
| 2 | .250 | 5.17 x 10^-5 |
| 3 | 0.314 | 6.49 x 10^-5 |
| 4 | 0.437 | 9.03 x 10^-5 |
| 5 | 0.521 | 1.08 x 10^-5 |
I am supposed to complete a table for each of them; table looks like this for all five of the mixtures:
| Mixture ## | [Fe^3+] | [SCN^-1] | [FeSCN^2+] |
| Initial | |||
| Change | |||
| Equilibrium |
Kc=
But I'm not sure how to get started; the data that we used was:
| 1 | 2 | 3 | 4 | 5 | |
| Volume Fe(NO3)3 | 5.00 | 5.00 | 5.00 | 5.00 | 5.00 |
| Volume KSCN solution (mL) | 1.00 | 2.00 | 3.00 | 4.00 | 5.00 |
| Volume H2O (mL) | 4.00 | 3.00 | 2.00 | 1.00 | 0 |
all of them are to equal 10mL (all mixtures)
I think this is all the information that is needed; Its fine if I only get shown at least like 2, I just want to make sure I understand it
Corrections- In 5th line it will be FeSCN2+ instead of FeSCN3+
From given data it is clear that we don't have any FeSCN2+ ions in our initial solution because we are adding only Fe(NO3)3 and KSCN. So the initial concentration of FeSCN2+ should be zero but it is given 1.50 x 10-4 M. I don't know how.
Well, I will solve this question for both cases if initial concentration is zero or if initial concentration is anyhow non zero. ( but the second case of non zero concentration is not logical. Please check it out again)
Case -1
change in concentration =
(equilibrium - initial)concentration similar calculations for all
other mixtures can be done.
Case - 2

LAB INFORMATION: Initial Fe3+ =2.00 x 10-3 M SCN- = 2.00 x 10-3 M FeSCN3+ =...
Lab Report: Determination of Kc for a Complex Ion Formation
tube
2.00e-3 Fe3+ (mL)
2.00E-3M SCN- (mL)
water (mL)
initial conc. Fe3+
initial conc. SCN-
1
5.00
5.00
0
1.00e-3M
1.00E-3M
2
5.00
4.00
1.00
1.00E-3M
8.00E-3M
3
5.00
3.00
2.00
1.00E-3,
6.00E-3M
4
5.00
2.00
3.00
1.00E-3M
4.00E-3M
5
5.00
1.00
4.00
1.00E-3M
2.00E-3M
10ml of 0.200M Fe3+, 2.00ml of 0.00200M SCN-, AND 8.00ml of
water results in an eq. [FeSCN2+] IN Standard
Soln.:2.00E-4M
Could you please explain how...
4. Calculate the initial concentrations of Fe3+ and SCN? in each of the equilibrium mixtures you will prepare using the information in Table 2 Volume of delonged water (ml) Total volume ml) Table 2. Preparation of Equilibrium Mixtures Volume of Volume of Volume of 0.10 M Mixture 0.00200 M 0.00200 M HNO, (ml) KSCN (ML) Fe(NO3)3 (ml) 1,50 3.00 1.00 2.00 1.00 2.50 1.00 1.50 350 1.00 1.00 2.50 2.00 4.50 4.50 4.50 4.50 4.50 10.00 10.00 10.00 10.00 10.00...
Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Volume in mL 2.00 x 103 M Fe(NO) Volume in mL, Depth in mm Volume in ml. 2.00 x 103 M Method I Mixture Unknówn KSCN Water Absorbance Standard FESCNP 4mL 1 5.00 x 10 M 1,00 .227 3mL 2 5,00 202 x 10 M 2,00 90 x 10 M .304 3 5,00 3.00 2mL 955 x 104 M I ImL 4 5.00 4,00 19x 10...
How do I calculate [SCN-]i given that:
Stock [Fe3+] = 0.00200 M
Stock [SCN-] = 0.00200 M
[Fe+3] [SCN]: of Table 2. Initial Data to Determine Keq Trial Volume Volume Total | Absorbance of Fe+3 volume KSCN(mm) (mm) (mL) hao 5.00 10.00 10.110 | 2 200 5.00 10.00 0.225 3 3.00 5.00 10.00 0.359 4 4.00 5.00 10.00 0.479 5 5.00 5.00 10.00 0.626 11.00×10 -3 2.00x104
1. A student mixes 5.00 mL of 2.00 x 10 M Fe(NO3)3 with 5.0 mL of 2.00 x 10-3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN+2 is 1.2 x 104 M. Find the Kc for the reaction of Fe (aq) + SCN (aq) → FeSCN2(aq) using the following steps. a. Find the initial concentration of Fe and SCN. (Use Equation 4). Record the value in the ICE Chart below. b. What is the equilibrium...
Consider the following: A student mixes 5.00 mL 2.00 × 10−3 M Fe(NO3)3 with 3.00 mL 2.00 × 10−3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.28 × 10−4 M. Find Kc for the reaction Fe3+(aq) + SCN−(aq) ↔ FeSCN2+(aq). a. What is the initial concentration of Fe3+ in the reaction mixture? [Fe3+] = ___ x 10-3 M b. What is the initial concentration of SCN- in the reaction mixture? [SCN-] = ___...
A student made solution #3 using the experimental method in this
lab, and measured an absorbance of 0.559. The starting reagents are
2.00 x 10-3 M Fe(NO3)3 and 2.00 x
10-3 M KSCN.
The amount of absorption is proportional to the concentration
of
FeSCN2+. This relationship – true for many solutions – is called
“Beer’s Law”, and has the simple equation:
A = bc
where “A” is the absorption, “b” is 5174.6 for FeSCN2+ and “c”
is molarity
Make Five...
Section Name Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 X 10M Fe(NO), with 5.00 ml 2.00 x 10-M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN is 1.40 x 10M. Find K for the reaction Fe(aq) + SCN (aq) FeSCN2(aq). Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3.) (5.00 x103 LX (300X163) = (x 103...
What's the concentration of [FeSCN2+] using limiting
reactant theory and equation?
For each test tube solution enter the initial concentration of Fe+ and SCNthe equilibrium concentration of FeSCN2: into the ICE table given. Complete entries for the rest of the table and calculate the K value for each of the tables. The values of K should be confined to a narrow range to reflect constancy. Comment on the quality of your work in this regard and calculate the average K....
1- A calibration plot to model Beer’s Law is constructed for a range of [FeSCN]2+ solutions between 1.00 x 10-5 M and 5 x 10-4 M. The equation of the line is y=5325x. Determine the molar absorptivity of FeSCN2+ in this range. 2- An unknown FeSCN2+ solution is prepared by adding 8.00 mL of 2.00 x 10-3 M Fe(NO3)3, 4.00 mL of 2.00 x 10-3 M KSCN, and 3.00 mL of water. The absorbance of the unknown [FeSCN]2+ solution is...