Question

3.0 mL of 0.02 M Fe(NO3)3 solution is mixed with 3.0 mL of 0.002 M NaNCS and diluted to the mark with HNO3 in 10 mL volumetric flask. The blood-red [Fe(NCS)]2+ ion that forms has an equilibrium molar concentration of 2.5*10-4 mol/L as determined from the calibration plot. Calculate the Kc for [Fe(NCS)]2+ formation. Assume the volumes are additive.

Answer 1

Answer:

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$.

Explanation:

$Moles=Concentration\times Volume (L)$

$Fe(NO_3)_3(aq)\rightarrow Fe^{3+}(aq)+3NO_3^{-}(aq)$

$[Fe(NO_3)_3]=0.02 M=[Fe^{3+}]$

Concentration of ferric ion = $[Fe^{3+}]=0.02 M$

Volume of ferric solution = 3.0 mL = 0.003 L

Moles of ferric ion  $=0.02 M\times 0.003 L$

1 mL = 0.001 L

$NaNCS(aq)\rightarrow Na^+(aq)+NCS^-(aq)$

$[NaNCS]=0.002 M=[NCS^-]$

Concentration of $NCS^-$ ion = $[NCS^{-}]=0.002 M$

Volume of $NCS^-$ ion solution = 3.0 mL = 0.003 L

Moles of $NCS^-$ ion= $0.002 M\times 0.003 L$

Volume of nitric acid solution = 10 mL = 0.010 L

After mixing all the solution the concentration of ferric ion and $NCS^-$ ion will change

Total volume of solution = 0.003 L + 0.003 L + 0.010 L = 0.016 L

Initial concentration of ferric ion before reaching equilibrium :

= $\frac{0.02 M\times 0.003 L}{0.016 L}=0.00375 M$

Initial concentration of $NCS^-$ ion before reaching equilibrium :

= $\frac{0.002 M\times 0.003 L}{0.016 L}=0.000375 M$

$Fe^{3+}+NCS^-\rightleftharpoons [Fe(NCS)]^{2+}$

Initially:

0.00375 M    0.000375 M       0

At equilibrium :

(0.00375-x)   (0.000375-x)       x

Equilibrium concentration of $[Fe(NCS)]^{2+}=x=2.5\times 10^{-4} M$

The expression of equilibrium constant for formation $[Fe(NCS)]^{2+}$ is given by :

$K_c=\frac{[[Fe(NCS)]^{2+}]}{[Fe^{3+}][NCS^-]}$

$K_c=\frac{x}{(0.00375-x)\times (0.000375-x)}$

$K_c=\frac{2.5\times 10^{-4} }{(0.00375-2.5\times 10^{-4})\times (0.000375-2.5\times 10^{-4})}$

$K_c=5.7\times 10^2$

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$.