A standard solution of FeSCN^2+ is prepared by combining 6.00 mL of 0.00150 M Fe

Question

A standard solution of FeSCN^2+ is prepared by combining 6.00 mL of 0.00150 M Fe(NO_3)_3, 1 065 g KSCN, and deionized water in a 250 mL volumetric flask. The net ionic equation for the reaction is below. Fe^3+ + SCN Fe(SCNp+ What allows us to assume that the reaction goes essentially to completion? The concentration of KSCN is much higher than that of Fe(NO_3)_3; The reaction quotient Q is greater than K_c. The equilibrium reaction has a very high K_c. Under the conditions given above, LeChatelier's principle dictates that the reaction shifts to the left. The concentration of Fe(NO_3)_3; is much higher than the concentration of KSCN. Based on that assumption, what is the equilibrium concentration of Fe(SCN)^2? For clarification see the example calculation on page 10-7 of your lab manual. Once checked, the calculated molarity will be copied to the appropriate space on Data Sheet page 10-11.

Explanation / Answer

number of moles of Fe(NO3)3 = 6 * 0.0015 = 0.009 m moles
concentration of Fe(NO3)3 solution = 0.009 / 250 = 3.6 * 10^-5
number of moles of KSCN = 1.06 / 97.2 = 1.091 8 10^-2 moles
Concentration of KSCN = 1.091 * 10^-2 /250 = 4.364 * 10^-5
so, number of moles of KSCN is higher. Hence for the completion of reaction we have to increase the number of moles of Fe(NO3)3 in the given solution.
so the concentration of Fe(NO3)3 is much higher the concnetration of KSCN

           Fe+3             +       SCN-           ---->     Fe(SCN)^2+
initial         3.6*10^-5              4.363 * 10^-5                  0
at equili (3.6 * 10^-5 - x)    (4.364*10^-5 - x)              x
Kc = [Fe(SCN)^2+]/[Fe+3][SCN-]
Kc = x/((3.6 * 10^-5 - x)*(4.364*10^-5 - x))
If we know Kc we can find x value and it is equilibrium concentration of Fe(SCN)^2+

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.