Question

1. fe³⁺(aq) + kscn(s) → fescn²⁺(aq) + k⁺(aq) to determine the moles of fe³⁺(aq) in a 100. ml sample of an unknown solution, excess kscn(s) is added to c fe³⁺(aq) into the dark red species fescn²⁺(aq), as represented by the equation above. the absorbance of fesc different concentrations is shown in the graph below. if the absorbance of the mixture is 0.20 at 453 nm, how many moles of fe³⁺(aq) were present in the 100. ml that any volume change due to adding the kscn(s) is negligible.) (a) 4×10⁻⁴ mol (b) 3×10⁻⁴ mol (c) 4×10⁻⁶ mol (d) 3×10⁻⁶ mol the diagram shown represents water molecules () as they interact with two ions, x and y, in solution.

Explanation:

Step1: Determine the concentration of \(FeSCN^{2 +}\)

From the graph, when the absorbance is \(0.20\), the concentration of \(FeSCN^{2+}\) is \(4\times10^{-5}M\).

Step2: Calculate the moles of \(Fe^{3+}\)

Since \(Fe^{3+}(aq)+KSCN(s)\to FeSCN^{2 +}(aq)+K^{+}(aq)\) (mole ratio \(1:1\) between \(Fe^{3+}\) and \(FeSCN^{2+}\)).
The volume of the solution \(V = 100\space mL=0.100\space L\).
Using the formula \(n = C\times V\), where \(C = 4\times 10^{-5}M\) and \(V=0.100\space L\).
\(n=(4\times 10^{-5}\space mol/L)\times0.100\space L = 4\times 10^{-6}\space mol\)

Answer:

C. \(4\times 10^{-6}\space mol\)