Question

A mixture is made by combining 10.9 mL of 4.1×10−3 M KSCN, 6.1 mL of 4.5×10−2 M Fe(NO3)3, and 12.5 mL of 0.484 M HNO3. Calculate concentraion of Fe3+

Answer #1

The reaction here is

Fe^{3+}(aq) + SCN^{–}(aq)
FeSCN^{2+}(aq)

[Fe^{3+}]eq = [Fe^{3+}]initial –
[FeSCN^{2+}]eq

[SCN^{–} ]eq = [SCN^{–} ]initial –
[FeSCN^{2+}]eq

10.9 mL of 4.1x10^{−3} M KSCN = (4.1x10^{−3} x
10.9)/1000 = 4.47 x 10^{-5} moles

6.1 mL of 4.5x10−2 M Fe(NO_{3})_{3}
=(4.5x10^{−2} x 6.1)/1000 = 2.74 x 10^{-4}
moles

Since SCN is less than Fe the amount of Fe in solution will be
the amount that does not react with SCN^{-}

which is 2.74 x 10^{-4} moles - 4.47 x 10^{-5}
moles = 2.30 x 10^{-4} moles

To get concentration we need to know the total volume which is 10.9 mL + 6.1 mL + 12.5 mL = 29.5 mL

2.30 x 10^{-4} moles in 29.5 mL will be
**7.79 x 10 ^{-3} M**

A mixture is made by combining 17.8 mL of 6.1×10−3 M KSCN, 9.6
mL of 9.5×10−2 M Fe(NO3)3, and 8 mL of 0.29 M HNO3. Calculate the
concentration of Fe3+ in the solution.

A mixture is made by combining 12.9 mL of 5.4×10−3 M KSCN, 8.3
mL of 2.2×10−2 M Fe(NO3)3, and 7.7 mL of 0.34 M HNO3. Calculate the
concentraion of Fe3+ in the solution.

A solution of 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300
M KSCN, and 3.00 mL of 1.0 M HNO3 is mixed together and allowed to
reach equilibrium. The concentration of Fe(SCN)2+ is found to be
2.72�10-4 M at equilibrium. Calculate the equilibrium constant for
this reaction.

A
solution of 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300 M
KSCN, and 3.00 mL of 1.0 M HNO3 is mixed together and allowed to
reach equilibrium. The concentration of Fe(SCN)^2+ is found to be
2.72 x 10^-4 M at equilibrium. Calculate the equilibrium constant
for this reaction.

1. A student mixes 5.00 mL 2.00 x 10-3 M Fe(NO3) in 1 M HNO3
with 3.00 mL 2.00 x 10-3 M KSCN and 2.00 mL of water. She finds
that in the equilibrium mixture the concentration of FeSCN2+ is 7.0
x 10-5 M. Find Kc for the reaction Fe3+ (aq0 + SCN- (aq0
------Fe(SCN)2+ (aq).
Step 1 Find the number of moles Fe3+ and SCN- initially
present.
Step 2. How many moles of FeSCN2+ are in the mixture at...

5.00 mL of 2.70E-3 M Fe(NO3)3 is mixed with 1.00 mL of 2.79E-3 M
KSCN and 4.00 mL of water. The equalibrium molarity of Fe(SCN)2+ is
found to be 6.00E-5 M. If the reaction proceeds as shown below,
what is the equilibrium molarity of Fe3+ and SCN-?

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 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...

A student mixes 5.00 mL of 2.00 x 10‐3 M Fe(NO3)3 with 5.00 mL
2.00 x 10‐3 M KSCN. She finds that in the equilibrium mixture the
concentration of FeSCN+2 is 1.40 x 10‐4 M.
a. What is the initial concentration in solution of
the Fe+3 and SCN‐ ?
b. What is the equilibrium constant for the
reaction?
c. What happened to the K+ and the NO3 ‐ ions in this
solution?

2.) FeSCN2+ equilibrium concentration was found to be 3.20 x
10-5 M in a solution made by mixing 5.00 mL of 1.00 x 10-3 M
Fe(NO3)3 with 5.00 mL of 1.00 x 10-3 M HSCN. The H+ concentration
is maintained at 0.500 M at all times since the HSCN and Fe3+
solutions were prepared using 0.500 M HNO3 in place of distilled
water. a. How many moles FeSCN2+ are present at equilibrium? b. How
many moles each of Fe3+ and...

In lab data:
Trial
Fe(No3)3 mL
KSCN mL
H2O ML
Total Vol. mL
%T
1
5.00
5.00
0
10.00
13.2
2
5.00
4.00
1.00
10.00
18.4
3
5.00
3.00
2.00
10.00
24.2
4
5.00
2.00
3.00
10.00
43.4
trial
%Transmittance
Absorption
[FeSCN] 2+
1
13.2
?
?
2
18.4
3
24.2
4
43.4
How do you calculate Absorption?
Conc. Fe(NO3)3 solution used ? Conc. KSCN solution used ?

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