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28 days ago

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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 f

lask. 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.

1 answer:

lorasvet [2.5K]28 days ago

5 0

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 ions = $[Fe^{3+}]=0.02 M$

Volume of the ferric solution is 3.0 mL or 0.003 L.

Calculating moles of ferric ions gives $=0.02 M\times 0.003 L$

1 mL = 0.001 L

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

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

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

Volume of the $NCS^-$ ion solution is also 3.0 mL or 0.003 L.

The moles of $NCS^-$ ions yields $0.002 M\times 0.003 L$

Volume of the nitric acid solution is 10 mL or 0.010 L.

Upon mixing, the concentrations of both ferric ions and $NCS^-$ ions will adjust.

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

Initial concentration of ferric ions before reaching equilibrium:

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

Initial concentration of $NCS^-$ ions before equilibrium is reached:

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

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

In the beginning:

0.00375 M 0.000375 M 0

At equilibrium:

(0.00375-x) (0.000375-x) x

Concentration of $[Fe(NCS)]^{2+}=x=2.5\times 10^{-4} M$ at equilibrium.

The expression for the equilibrium constant regarding the formation of $[Fe(NCS)]^{2+}$ is formulated as:

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

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

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

$K_c=5.7\times 10^2$

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

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