# Question c450e

Sep 25, 2014

The equilibrium constant is 3 × 10^2.

#### Explanation:

First, write the balanced chemical equation with an ICE table.

$\textcolor{w h i t e}{m m m m m m m m m m} \text{Fe"^"3+" color(white)(m)+ color(white)(m)"SCN"^"-"color(white)(m) ⇌color(white)(m) "FeSCN"^"2+}$
$\text{I/mol·L"^"-1":color(white)(mmm)1.0 × 10^"-3"color(white)(mm) 8.0 × 10^"-4} \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmmml) "-"xcolor(white)(mmmmmll) "-"xcolor(white)(mmmmmll) "+} x$
$\text{E/mol·L"^"-1": color(white)(l)1.0 × 10^"-3" - xcolor(white)(m) 8.0 × 10^"-4} - x \textcolor{w h i t e}{m m m} x$

At equilibrium, ["FeSCN"^"2+"] = 1.7 × 10^"-4"color(white)(l) "mol/L" = x color(white)(l)"mol/L"

So x = 1.7 × 10^"-4"

Then

["Fe"^"3+"] = (1.0 × 10^"-3" - x)color(white)(l) "mol/L" = (1.0×10^"-3" – 1.7×10^"-4")color(white)(l) "mol/L"

= 8.3 × 10^"-4" color(white)(l)"mol/L" (1 significant figure + 1 guard digit)

and

["SCN"^"-"] = (8.0 × 10^"-4" - x)color(white)(l) "mol/L" = (8.0×10^"-4" - 1.7×10^"-4") color(white)(l)"mol/L"

= 6.3 × 10^"-4" color(white)(l)"mol/L"

K_"c" = (["FeSCN"^"2+"])/(["Fe"^"3+"] ["SCN"^"-"]) = (1.7 × 10^"-4")/( 8.3 × 10^-"4 "× 6.3 × 10^-4) = 3 × 10^2#

Note: The answer can have only 1 significant figure, because the initial concentration of $\text{Fe"^"3+}$ has only one digit after the decimal point. If you need more precision, you will have to recalculate.