Question

In the calibration part of the experiment (Part 1), the reactants are combined with one as a limiting reagent. According to the equation Fe3+(aq) + SCN-(aq) ⇌ Fe(SCN)]2+(aq), if 1.0 mL of 0.2 M Fe3+(aq), 1.0 mL of 0.002 M SCN-(aq) and 69.0 mL of H2O are mixed together as the initial solution, what will be the equilibrium concentration of SCN-(aq)?

Answer 1

Fe³⁺(aq) + SCN^-(aq) <-----> Fe(SCN)]²⁺(aq)
Initial no of moles of Fe³⁺ = 0.2 mol/L * ((1+69)/1000))L = 0.014 mol
Initial no of moles of SCN^- = 0.002mol/L * ((1+69)/1000)L = 1.4*10^-4mol
initial no of mol of Fe(SCN)]²⁺ = 0 mol
Assume that you have X mol of each reactant reacted
So equilibrium concentrations will be,
[Fe³⁺] = 0.014-X ; [SCN^-] = 1.4*10^-4 - X ; [Fe(SCN)]²⁺] = X
K_c (not given in question) = X / (0.014-X) (1.4*10^-4 - X)
Substitute K_c and solve for X from the above expression.
and then put the X value in ( 1.4*10^-4 - X ) to get equilibrium concentration of [SCN^-]