Question #c450e

1 Answer
Sep 25, 2014

Answer:

The equilibrium constant is #3 × 10^2#.

Explanation:

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

#color(white)(mmmmmmmmmm)"Fe"^"3+" color(white)(m)+ color(white)(m)"SCN"^"-"color(white)(m) ⇌color(white)(m) "FeSCN"^"2+"#
#"I/mol·L"^"-1":color(white)(mmm)1.0 × 10^"-3"color(white)(mm) 8.0 × 10^"-4"color(white)(mmmm) 0#
#"C/mol·L"^"-1":color(white)(mmmml) "-"xcolor(white)(mmmmmll) "-"xcolor(white)(mmmmmll) "+"x#
#"E/mol·L"^"-1": color(white)(l)1.0 × 10^"-3" - xcolor(white)(m) 8.0 × 10^"-4" - xcolor(white)(mmm) 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 #"Fe"^"3+"# has only one digit after the decimal point. If you need more precision, you will have to recalculate.