In part 3, it is assumed that the small concentration of SCN^- in the pre-equili

Question

In part 3, it is assumed that the small concentration of SCN^- in the pre-equilibrium mixture completely reacts to form FeSCN^2+ at equilibrium. Which ion, Fe^3+ or SCN^-, is present in excess? Fe^3+ SCN^- Rearrange the K_eq expression to give the equilibrium ratio of [FeSCN^2+] to [SCN^-] in terms of K_eq and [Fe^3+]. In other words, express [FeSCN^2+]/[SCN^-] in terms of K_eq and [Fe^3+]. Calculate the [SCN^-] in solution A, where 5.3 mL of 1.5 Times 10^-3 M NaSCN is diluted to 46.0 mL with 0.1 M Fe(NO_3)_3. (Include units in your answer. More information.) Calculate the [FeSCN^2+] in the solutions prepared by diluting volumes of solution A in part c with 0.1 M Fe(NO_3)_3 to a final volume of 1.0 Times 10^1 mL. (Use the rounded value you calculated for part (c) to calculate your answers to part (d). Include units in your answer. More information.)

Explanation / Answer

Fe+3 + SCN- ------> FeSCN+2
since SCN- is low concentration measns then excess ion concentration is Fe+3
Keq = [FeSCN+2]/[Fe+3][SCN-]
Keq*[Fe+3] = [FeSCN+2]/[SCN-]
number of moles of SCN- = 5.3 * 1.5 * 10^-3 = 0.00795
number of moles of Fe(NO3)3 = 46 * 0.1 = 4.6
so SCN- is completely utilized in the reaction since it is present in small amounts so it is called limiting reagent.
concentration of SCN- = 0.00795/(5.3+46) = 1.5497 * 10^-4

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