Question

am confused how to start this.

for my lab i obtained 24ml of 0.000075M Fe(NO3)3, 40ml of 1M KSCN and 45ml 0f 0.1M HNO3.

i was asked to pipet different volumes into 12 testubes.
testtube 1....4.00ml Fe(NO3)3, 3.00ml KSCN and 1.00ml HNO3.

Calculate the molarity of the complex [Fe(SCN)]^2+ that was present at equilibrium.

Am confusing my ideas please help

Answer 1

The total no. of millimoles of Fe³⁺ (i.e. Fe(NO₃)₃) = 24 mL * 0.000075 mmol/mL = 0.0018 mmol

The total no. of millimoles of SCN^- (i.e. KSCN) = 40 mL * 1 mmol/mL = 40 mmol

In test tube 1,

The no. of millimoles of Fe³⁺ = 4 mL * 0.000075 mmol/mL = 0.0003 mmol

The no. of millimoles of SCN^- = 3 mL * 1 mmol/mL = 3 mmol

The volume of HNO₃ = 1 mL

Therefore, the total volume of the solution = 4 + 3 + 1 = 8 mL

Hence, the molariy of the complex [Fe(SCN)]²⁺, which is formed by the 1:1 combination of Fe³⁺ and SCN^- = 0.0003 mmol/8 mL ~ 3.75*10^-5 M

The value is not exactly, 3.75*10^-5 M, but it is approximately 3.75*10^-5 M. because at equilibrium, some [Fe³⁺] should also be present.