1. A student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO3)3 with 5.00 mL of 2.00 x 10-3
ID: 890176 • Letter: 1
Question
1. A student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO3)3 with 5.00 mL of 2.00 x 10-3 M KSCN. She finds that in the equilibrium mixture the of concentration of FeSCN2+ is 1.40 x 10-4 M.
a) What is the initial concentration in solution of the Fe3+ and SCN- ?
b) What is the equilibrium constant for the reaction?
2. Assume that the reaction studied is actually: Fe3+ (aq) + 2 SCN- (aq) Fe(SCN)2+ (aq)
a) What is the equation for determining the equilibrium constant?
b) Using the information from question 1 and assuming [Fe(SCN)2+] = 1.40 x 10-4 M, calculate the equilibrium concentration of Fe3+ and SCN- .
c) Determine the numerical value of K.
Explanation / Answer
a)
Total volume = 5+5 =10 mL = 1*10^-2 L
Number of moles of Fe3+ = M*V
=2*10^-3 M * 5*10^-3 L
= 1*10^-5 mol
[Fe3+] = number of moles / total volume
= (1*10^-5) / (1*10^-2)
= 1*10^-3 M
Number of moles of SCN- = M*V
=2*10^-3 M * 5*10^-3 L
= 1*10^-5 mol
[SCN-] = number of moles / total volume
= (1*10^-5) / (1*10^-2)
= 1*10^-3 M
b)
Fe(NO3)3 + KSCN ---> Fe (SCN) 2+ + KNO3
1*10^-3 1*10^-3 0 0 <----initial
1*10^-3-x 1*10^-3-3x x x <---- at equilibrium
Given that at equilibrium:
[FeSCN 2+] is 1.4*10^-4 M
so,
x= 1.4*10^-4 M
Kc = [FeSCN 2+] [KNO3] /{[Fe(NO3)3] [KSCN]}
= x*x / {(1*10^-3 - x) (1*10^-3 - x)}
= x^2 / (1*10^-3 - x)^2
= (1.4*10^-4)^2 / (1*10^-3 - 1.4*10^-4)^2
=0.0264
Answer: 0.0264
I am allowed to answer only 1 question at a time. Please ask other as different question
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.