In this experiment you will be adding Fe(NO3)3·9H2O to KSCN to form a bright ora

Question

In this experiment you will be adding Fe(NO3)3·9H2O to KSCN to form a bright orange complex, [Fe(SCN) 2, according to the following balanced equation: Fe"(aq) + SCN'(aq) [Fe(SCN) 2 (aq) The resulting equilibrium expression is: 0o You are given the following information: [Fe(NO3h-9H2O)stock, 0.00350 M [KSCNJstock: 0.000700 M o Analytical wavelength: 470 nm You make up the solutions to the following specifications: s 00020mol Fe(NO)3.9H20 KSCN 12.00 10.00 8.00 6.00 4.00 Vial umber Initial Volume, mL Initial Volume, mLFinal Volume, mL. 4.00 6.00 8.00 10.00 12.00 = (4.00 + 12.00-16.00 = 16.00 = 16.00 = 16.00 = 16.00 4 119

Explanation / Answer

The given concentration of the stock solutions are

Concentration of [Fe(NO3)3].9H2O = concentration of Fe3+, [Fe3+] = 0.00350 M

Concentration of KSCN = concentration of SCN-, [SCN-] = 0.000700 M

Vial-1: Volume of [Fe(NO3)3].9H2O taken = 4.00 mL = 4.00x10-3 L

Hence moles of Fe3+ in vial -1 = MxV = 0.00350 M x 4.00x10-3 L = 1.4x10-5 mol

Volume of KSCN taken = 12.00 mL = 12.00x10-3 L

Hence moles of SCN- in vial -1 = MxV = 0.000700 M x 12.00x10-3 L = 8.4x10-6 mol

Final volume of Vial-1, Vf = 4.00 mL + 12.00 mL = 16.0 mL = 16.00x10-3 L

Hence concentration of Fe3+ in vial -1, [Fe3+] = moles of Fe3+ / Vf = 1.4x10-5 mol / 16.00x10-3 L

= 8.75x10-4 M(answer)

concentration of SCN- in vial -1, [SCN-] = moles of SCN- / Vf = 8.4x10-6 mol / 16.00x10-3 L

= 5.25x10-4 M(answer)

Vial-2: Volume of [Fe(NO3)3].9H2O taken = 6.00 mL = 6.00x10-3 L

Hence moles of Fe3+ in vial -2 = MxV = 0.00350 M x 6.00x10-3 L = 2.1x10-5 mol

Volume of KSCN taken = 10.00 mL = 12.00x10-3 L

Hence moles of SCN- in vial -2 = MxV = 0.000700 M x 10.00x10-3 L = 7.0x10-6 mol

Final volume of Vial-2, Vf = 6.00 mL + 10.00 mL = 16.0 mL = 16.00x10-3 L

Hence concentration of Fe3+ in vial -2, [Fe3+] = moles of Fe3+ / Vf = 2.1x10-5 mol / 16.00x10-3 L

= 1.31x10-3 M(answer)

concentration of SCN- in vial -2, [SCN-] = moles of SCN- / Vf = 7.0x10-6 mol / 16.00x10-3 L

= 4.38x10-4 M(answer)

Vial-3: Volume of [Fe(NO3)3].9H2O taken = 8.00 mL = 8.00x10-3 L

Hence moles of Fe3+ in vial -3 = MxV = 0.00350 M x 8.00x10-3 L = 2.8x10-5 mol

Volume of KSCN taken = 8.00 mL = 8.00x10-3 L

Hence moles of SCN- in vial -3 = MxV = 0.000700 M x 8.00x10-3 L = 5.6x10-6 mol

Final volume of Vial-3, Vf = 8.00 mL + 8.00 mL = 16.0 mL = 16.00x10-3 L

Hence concentration of Fe3+ in vial -3, [Fe3+] = moles of Fe3+ / Vf = 2.8x10-5 mol / 16.00x10-3 L

= 1.75x10-3 M(answer)

concentration of SCN- in vial -3, [SCN-] = moles of SCN- / Vf = 5.6x10-6 mol / 16.00x10-3 L

= 3.5x10-4 M(answer)

Vial-4: Volume of [Fe(NO3)3].9H2O taken = 10.00 mL = 10.00x10-3 L

Hence moles of Fe3+ in vial -4 = MxV = 0.00350 M x 10.00x10-3 L = 3.5x10-5 mol

Volume of KSCN taken = 6.00 mL = 6.00x10-3 L

Hence moles of SCN- in vial -4 = MxV = 0.000700 M x 6.00x10-3 L = 4.2x10-6 mol

Final volume of Vial-4, Vf = 10.00 mL + 6.00 mL = 16.0 mL = 16.00x10-3 L

Hence concentration of Fe3+ in vial -4, [Fe3+] = moles of Fe3+ / Vf = 3.5x10-5 mol / 16.00x10-3 L

= 2.19x10-3 M(answer)

concentration of SCN- in vial -3, [SCN-] = moles of SCN- / Vf = 4.2x10-6 mol / 16.00x10-3 L

= 2.63x10-4 M(answer)

Vial-5: Volume of [Fe(NO3)3].9H2O taken = 12.00 mL = 12.00x10-3 L

Hence moles of Fe3+ in vial -5 = MxV = 0.00350 M x 12.00x10-3 L = 4.2x10-5 mol

Volume of KSCN taken = 4.00 mL = 4.00x10-3 L

Hence moles of SCN- in vial -5 = MxV = 0.000700 M x 4.00x10-3 L = 2.8x10-6 mol

Final volume of Vial-4, Vf = 12.00 mL + 4.00 mL = 16.0 mL = 16.00x10-3 L

Hence concentration of Fe3+ in vial -5, [Fe3+] = moles of Fe3+ / Vf = 4.2x10-5 mol / 16.00x10-3 L

= 2.63x10-3 M(answer)

concentration of SCN- in vial -3, [SCN-] = moles of SCN- / Vf = 2.8x10-6 mol / 16.00x10-3 L

= 1.75x10-4 M(answer)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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