A 2.50ml aliquot of 0.0103 M Fe(NO3)3 is added to 2.50ml of 0.00101 M NaSCN and the following equilibrium is reached Fe(3+) + SCN(-) =FeNCS(2+). The concentration of FeNCS(2+) was determined to be 3.50 multiplied by 10(-4) M at equilibrium. Calculate Kc?

Question

Henry Antrim · Accepted Answer

Thus, iron (III) nitrate and sodium thiocyanate have the following general reaction (all species are in aqueous solution)

#Fe(NO_3)_3 + NaSCN rightleftharpoons FeSCN(NO_3)_2 + NaNO_3#

LIke you know, the net ionic reaction shows the equilibrium between the ferric ion, #Fe^(3+)#, and the thiocyanate ion, #SCN^(-)#, with the red-orange colored #FeSCN^(2+)# complex ion

#Fe^(3+) + SCN^(-) rightleftharpoons FeSCN^(2+)#

Now that you are aware of the complex ion's equilibrium concentration, the following will help you solve the problem with less effort.

You will notice that the final volume of the solution will be twice the initial volumes because you have equal volumes of sodium thiocyanate and iron (III) nitrate mixed together.

For both compounds, the final concentrations in solution will be half of the initial concentration if the final volume doubles, which brings us to our next point about dilution calculations.

#[Fe(NO_3)_3] = [Fe^(3+)] = "0.0103 M"/2 = "0.00515 M"#, and

#[NaSCN] = [SCN^(-)] = "0.00101 M"/2 = "0.000505 M"#

Now think of the formation of the complex ion. Initially, you had no complex ion formed in solution. Then, at equilibrium, the concentration is given to be #3.50 * 10^(-4)"M"#. This means that the concentration of the complex ion went from zero to #3.50 * 10^(-4)"M"#.

Automatically, the concentrations of the ferric and thiocyanate ions must have decreased by the same amount. This happens because of the stochiometry of the reaction - 1 mole of #Fe^(3+)# reacts with 1 mole of #SCN^(-)# to produce 1 mole of #FeSCN^(2+)#.

Thus, the two ions' equilibrium concentrations are

#[Fe]^(3+) = "0.00515 M" - "0.000350 M" = "0.0048 M"#, and

#[SCN^(-)] = "0.000505 M" - "0.000350 M" = "0.000155 M"#

Simply enter these values into the equilibrium constant expression now.

#K_c = ([FeSCN^(2+)])/([Fe^(3+)] * [SCN^(-)]) = (0.000350)/(0.0048 * 0.000155) = 470.4#

Therefore, #K_c = 470# - rounded to three sig figs.

Jose Ayala · Answer

undefined To calculate the equilibrium constant $ K_c $, first, determine the initial concentrations of Fe(3+), SCN(-), and FeNCS(2+). Then, use the equilibrium concentrations to calculate $ K_c $ using the formula:

$$ K_c = \frac{{[\text{FeNCS}^{2+}]}}{{[\text{Fe}^{3+}][\text{SCN}^-]}} $$

Given:
- Initial concentration of Fe(NO3)3 = 0.0103 M
- Initial concentration of NaSCN = 0.00101 M
- Volume of both solutions = 2.50 mL

Calculate the moles of Fe(3+) and SCN(-) using the initial concentrations and volumes. Then, use the stoichiometry of the reaction to determine the change in concentration of each species at equilibrium. Finally, plug these equilibrium concentrations into the equilibrium constant expression to find $ K_c $.