A reaction mixture of 4.0 mL of 0.002 M SCN- and 5.0 mL of 0.002 M Fe3+ is diluted to 10.0 mL with 0.1 M HNO3 to form the blood-red FeNCS2+ complex. The equilibrium molar concentration of the FeNCS2+ determined from a standardization curve is 1.5x10^-4 mol/L. Calculate, in sequence, each of the following quantities in the aqueous solution to determine the equilibrium constant for the reaction:
Fe3+(aq) + SCN-(aq) <--> FeNCS2+(aq)
A) moles of FeNCS2+ that form in reaching equilibrium
B) moles of Fe3+ that react to form the FeNCS2+ at equilibrium
C) moles of SCN- that react to form the FeNCS2+ at equilibrium
D) moles of Fe3+ initially placed in the reaction system
e) moles of SCN- initially placed un the reaction system
f) moles of FE3+ (unreacted) at equilibrium
G) moles of SCN- (unreacted) at equilibrium
h) molar concentration of Fe3+ (unreacted) at equilibrium
i) molar concentration of SCN- (unreacted) at equilibrium
j) molar concentration of FeNCS2+ at equilibrium
k) Kc = [FeNCS2+]/([Fe3+][SCN-])
I'll get you started.
mols SCN^- initially = M x L = ?
mols Fe^3+ initially = M x L = ?
(SCN^-) initially = 0.002 x 4/10 = 8E-4M
(Fe^3+) initially = 0.002 x 5/10 = 0.001M
.........Fe^3+ + SCN^- ==> FeNCS^2+
I.......8E-4.....1E-3........0
C........-x.......-x.........x
E......8E-4-x...1E-4-x......1.5E-4
So you know x is 1.5E-4 and that allows you to calculate values for (Fe^3+) and SCN^-) at equilibrium and that allows you to calculate Kc.
The rest of that question is busy work but when you want to answer a mol question, it is mols = M x L = ?
To determine the quantities in the aqueous solution and the equilibrium constant for the reaction, let's go through the steps one by one:
A) moles of FeNCS2+ that form in reaching equilibrium:
The equilibrium molar concentration of FeNCS2+ is given as 1.5x10^-4 mol/L. Since the final volume is 10.0 mL, the moles of FeNCS2+ that form can be calculated using the equation:
moles = concentration x volume
moles of FeNCS2+ = 1.5x10^-4 mol/L x 10.0 mL
moles of FeNCS2+ = 1.5x10^-4 mol
B) moles of Fe3+ that react to form the FeNCS2+ at equilibrium:
From the balanced equation, the stoichiometric ratio between Fe3+ and FeNCS2+ is 1:1. Therefore, the moles of Fe3+ that react can be calculated as:
moles of Fe3+ = moles of FeNCS2+
moles of Fe3+ = 1.5x10^-4 mol
C) moles of SCN- that react to form the FeNCS2+ at equilibrium:
Again, by stoichiometry, the moles of SCN- that react can be determined as:
moles of SCN- = moles of FeNCS2+
moles of SCN- = 1.5x10^-4 mol
D) moles of Fe3+ initially placed in the reaction system:
The initial volume of Fe3+ is 5.0 mL, and the initial concentration is 0.002 M. Similar to step A, we can calculate the moles of Fe3+:
moles = concentration x volume
moles of Fe3+ = 0.002 M x 5.0 mL
moles of Fe3+ = 1.0x10^-5 mol
E) moles of SCN- initially placed in the reaction system:
Similarly, the initial volume of SCN- is 4.0 mL, and the initial concentration is 0.002 M. Therefore, the moles of SCN- can be calculated as:
moles of SCN- = 0.002 M x 4.0 mL
moles of SCN- = 8.0x10^-6 mol
F) moles of Fe3+ (unreacted) at equilibrium:
Since the moles of Fe3+ that reacted to form FeNCS2+ at equilibrium is the same as the moles of FeNCS2+ that form, the moles of Fe3+ (unreacted) at equilibrium is 0 moles.
G) moles of SCN- (unreacted) at equilibrium:
Similar to F, since the moles of SCN- that reacted to form FeNCS2+ at equilibrium is the same as the moles of FeNCS2+ that form, the moles of SCN- (unreacted) at equilibrium is also 0 moles.
H) molar concentration of Fe3+ (unreacted) at equilibrium:
The final volume of the solution is 10.0 mL, and there are no moles of Fe3+ (unreacted) at equilibrium. Thus, the molar concentration is 0 M.
I) molar concentration of SCN- (unreacted) at equilibrium:
Similar to H, the molar concentration of SCN- (unreacted) at equilibrium is also 0 M.
J) molar concentration of FeNCS2+ at equilibrium:
The equilibrium molar concentration of FeNCS2+ is given as 1.5x10^-4 mol/L.
K) Kc = [FeNCS2+]/([Fe3+][SCN-]):
Using the concentrations at equilibrium, we have:
Kc = ([FeNCS2+]) / ([Fe3+][SCN-])
Kc = (1.5x10^-4 mol/L) / (0 M x 0 M)
Since the molar concentrations of Fe3+ and SCN- (unreacted) at equilibrium are 0 M, the equilibrium constant Kc is undefined or infinite.
To solve this problem, we need to use the given information and apply the principles of equilibrium and stoichiometry. Let's go through each of the steps to calculate the quantities and determine the equilibrium constant, Kc.
A) Moles of FeNCS2+ that form in reaching equilibrium:
Using the molar concentration of FeNCS2+ at equilibrium (given as 1.5x10^-4 mol/L) and the total volume of the solution (10.0 mL or 0.01 L), we can calculate the moles of FeNCS2+ formed:
moles of FeNCS2+ = molar concentration x volume
moles of FeNCS2+ = 1.5x10^-4 mol/L x 0.01 L
B) Moles of Fe3+ that react to form FeNCS2+ at equilibrium:
Since the reaction is 1:1 between Fe3+ and FeNCS2+ (as shown in the balanced equation), the moles of Fe3+ that react is equal to the moles of FeNCS2+ formed.
C) Moles of SCN- that react to form FeNCS2+ at equilibrium:
Since the reaction is 1:1 between SCN- and FeNCS2+ (as shown in the balanced equation), the moles of SCN- that react is equal to the moles of FeNCS2+ formed.
D) Moles of Fe3+ initially placed in the reaction system:
To find the moles of Fe3+ initially placed, we need to calculate it based on the initial volume and the molar concentration of Fe3+.
moles of Fe3+ = molar concentration x volume
moles of Fe3+ = 0.002 mol/L x 0.005 L
E) Moles of SCN- initially placed in the reaction system:
To find the moles of SCN- initially placed, we need to calculate it based on the initial volume and the molar concentration of SCN-.
moles of SCN- = molar concentration x volume
moles of SCN- = 0.002 mol/L x 0.004 L
F) Moles of Fe3+ (unreacted) at equilibrium:
The moles of Fe3+ that remain unreacted at equilibrium can be calculated by subtracting the moles of Fe3+ that react (derived from part B) from the moles of Fe3+ initially placed (derived from part D).
G) Moles of SCN- (unreacted) at equilibrium:
The moles of SCN- that remain unreacted at equilibrium can be calculated by subtracting the moles of SCN- that react (derived from part C) from the moles of SCN initially placed (derived from part E).
H) Molar concentration of Fe3+ (unreacted) at equilibrium:
The molar concentration of Fe3+ (unreacted) at equilibrium can be calculated by dividing the moles of Fe3+ (unreacted) at equilibrium (derived from part F) by the total volume of the solution (10.0 mL or 0.01 L).
I) Molar concentration of SCN- (unreacted) at equilibrium:
The molar concentration of SCN- (unreacted) at equilibrium can be calculated by dividing the moles of SCN- (unreacted) at equilibrium (derived from part G) by the total volume of the solution (10.0 mL or 0.01 L).
J) Molar concentration of FeNCS2+ at equilibrium:
The molar concentration of FeNCS2+ at equilibrium is given as 1.5x10^-4 mol/L.
K) Kc = [FeNCS2+]/([Fe3+][SCN-]):
Kc is the equilibrium constant and can be calculated by dividing the molar concentration of FeNCS2+ at equilibrium (derived from part J) by the product of the molar concentrations of Fe3+ (unreacted) at equilibrium (derived from part H) and SCN- (unreacted) at equilibrium (derived from part I).
Please note that the exact calculations need to be performed using the given values and calculator to arrive at the specific numerical answers for each part.