A student mixes 5.00mL of 2.00x10^-3M Fe(NO3)3, 4.00mL of 2.00x10^-3 KSCN, and 1

Question

A student mixes 5.00mL of 2.00x10^-3M Fe(NO3)3, 4.00mL of 2.00x10^-3 KSCN, and 1.00mL of distilled water and finds that in the equilibrium mixture the concentration of FeSCN^2+ is 1.31x10^-4. (need to know how to do them ann the answers, so if I could see work being done of the problem to get the answer that would be awesome)

Fe^3+(aq) + SCN-(aq)---> FeSCN^2+(aq)

1. Find the number of moles of Fe^3+ and SCN- initially present.

2. How many moles of FeSCN^2+ are in the mixture at equilibrium?

3. How many moles of Fe^3+ and SCN- are used up in making the FeSCN^2+?

4. How many moles of Fe^3+ and SCN- remain in the solution at equilibrium?

5. What are the concentration of Fe^3+, SCN-, and FeSCN^2+ at equilibrium?

6. What is the value of Kc for this reaction?

Explanation / Answer

(1)
5.00x10^-3 L x 2.00x10^-3 M = 1.00x10^-5 mole Fe+3
4.00x10^-3 L x 2.00x10^-3 M = 8.00x10^-6 mole SCN-
(2)
10.00x10^-3 L x 1.31x10^-4 M = 1.131x10^-6 mol FeSCN+2
(3)
1.31x10^-6 mole each of Fe+3 and SCN- reacted to form FeSCN+2
(4)
1.00x10^-5 - 1.31x10^-6 = 8.69x10^-6 mole Fe+3 remains
8.00x10^-6 - 1.31x10^-6 = 6.69 mole SCN- remains
(5)
Fe+3 = 8.69x10^-6 mol / 1.0x10^-2 L = 8.69x10^-4 M
SCN- = 6.69x10^-6 mol / 1.0x10^-2 = 6.69x10^-4 M
FeSCN+2 = 1.31x10^-4 M
(6)
K = 1.31x10^-4 / (6.69x10^-4 x 8.69x10^-4) = 2.25x10^2

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