Incorrect 0/2 pts. Question 4 During lab, you used a Mohr pipet to add the follo
ID: 508474 • Letter: I
Question
Incorrect 0/2 pts. Question 4 During lab, you used a Mohr pipet to add the following volumes into a 25 mL volumetric flask, then followed the directions in the manual to make the stock solution. 0.200 M 0.200 0.00200 M 0.00200 M Fe(NO3 3 (initial M Fe(NO33 (final KSCN (nitial KSCN (final volume) volume) volume) volume) Stock 0.03 mL 4.77 mL. 9.69 mL 2.62 mL Solution Using the stock solution above, you made additional dilutions, as described below. Calculate the [FescN2 in Standard 1. Stock solution (initial Stock solution (final Water (initial Water (final Solution volume) volume) volume) volume 2.62 mL. 4.07 mL 1.86 mL Standard 1 9.27 mL Standard 2 5.00 mL 8.03 mL 3.63 mL 9.11 mL Standard 5.47 mL 2.79 mL 8.13 mL, 1.66 mL Standard 4 7.27 mL 0.56 mL. 8.01 mL 9.14 mL. Note: For8rading purposes, report your answer in units of mM. 228600.0000 Your final answer should be between 0.01 and 0.2 mM.Explanation / Answer
Answer for question 4:
According to the experiment one equivalent each of Fe(NO3)3 and KSCN is used in the reaction to form ferrous thiocyanate complex as shown in the equation below.
Fe(NO3)3 + KSCN <==============> [FeSCN)]2+
Stock solution of Fe(NO3)3:
Volume of 0.2 M Fe(NO3)3 used as per Mohr pipet: 9.69 ml – 0.03 ml = 9.66 ml
Number of mmoles of Fe(NO3)3 used = 0.2 x 9.66 mL = 1.932 mmoles
Volume of 0.002 M KSCN used as per Mohr pipet: 4.77 ml – 2.62 ml = 2.15 ml
Number of mmoles of KSCN used = 0.002 x 2.15 mL = 0.0043 mmoles
Total volume of stock solution = volume of Fe(NO3)3 used + volume of KSCN used
= 9.66 ml + 2.15 ml = 11.81 ml
As per above equation 1 eq. Fe(NO3)3 reacts with 1 eq. of KSCN to form [FeSCN]2+. Hence number of mmoles of [FeSCN]2+ formed will be equal to that of KSCN used since it is the limiting reagent.
Therefore number of mmoles of [FeSCN]2+ formed = 0.0043 mmoles in 11.81 ml stock solution
For preparing Standard solution 1:
Amount of stock solution used: 9.27 ml – 2.62 ml = 6.65 ml
Amount of water used: 4.07 ml – 1.86 ml = 2.21 ml
Total volume of Standard solution1: stock solution + water = 6.65 ml + 2.21 ml = 8.86 ml
Amount of [FeSCN]2+ in standard solution1:
11.81 ml of stock solution has 0.0043 mmoles [FeSCN]2+
Number of mmoles of [FeSCN]2+ present in 6.65 ml of stock solution:
Number of mmoles of [FeSCN]2+ = 6.65 ml x 0.0043 mmoles / 11.81 ml
= 0.002421 mmoles
Number of mmoles of [FeSCN]2+ = 0.002421 mM
(The answer will not be as mentioned in the question:
Expected range of number of mmoles of [FeSCN]2+: 0.01 - 0.2 mM.
Because the stock solution itself is expected to have a total of 0.0043mM which is nearly 50 times lower than the 0.2 mM expected value.)