Hi, if anyone could help, it would be greatly appreciated! Thanks!! To obtain [F

Question

Hi, if anyone could help, it would be greatly appreciated! Thanks!!

To obtain [FeSCN2+]eq in tube 5, you made the assumption that 100% of the ions SCN– had

reacted. Now that you know the value for Kc (average in table 5), build an ICE diagram to calculate the true percentage of SCN– reacted in tube 5. Comment on the validity of the assumption. (Make sure you include the ICE diagram in your anwer)

Absorbance Initial cocentrations [SCN lo (au) (M) 180 1.88x10A-4 1.70 10A4 0.680 0.001 3.80x10 -4. 1.15x10 -4 0.228 0.406 3.08x10 -4 0.001 5.70x10 A-4 0.001 7.60x10A-4 5.85x10 -4 0.577 4 0.707 0.001 9.50x10A-4 8.95x10 -4. K Concentrations at equilibrium [Fes le (MTI) (M) (M) 10A-4 2.30x10A-4 5.64x10 6.92x10 A-4 2.62x10 -4 1.69x10 3 4.15x 10A-4 1.75x10 -4 8.06x10A3 1.05x10 -4 5.50x10 -5 1.55x10 5 Average 4.13x10A4

Explanation / Answer

Fe^3+ + SCN- <==> [Fe(SCN)^2+]eq

Kc = [Fe(SCN)]^2+/[Fe^3+][SCN^-]

4.13*10^4 = x/(0.18-x)(1.88*10^-4-x)

or, x = 1.87 *10^-4

amount reacted = 1.87 *10^-4

% reacted = (1.87 *10^-4/1.88 *10^-4) *100% = 99.47 %

As the unreacted amount is very small, this assumption can be made.

Fe^3+ SCN^- Fe(SCN)^2+ initial 0.18 1.88*10^-4 0 change -x -x +x equilibrium 0.18-x 1.88*10^-4-x x

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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