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+" m m m m m m m m m m Fe 3+ m + m SCN - m ⇌ 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 I/mol⋅L -1 : m m m 1.0 × 10 -3 m m 8.0 × 10 -4 m m m m 0 "C/mol·L"^"-1":color(white)(mmmml) "-"xcolor(white)(mmmmmll) "-"xcolor(white)(mmmmmll) "+"x C/mol⋅L -1 : m m m m l - x m m m m m l l - x m m m m m l l + x "E/mol·L"^"-1": color(white)(l)1.0 × 10^"-3" - xcolor(white)(m) 8.0 × 10^"-4" - xcolor(white)(mmm) x E/mol⋅L -1 : l 1.0 × 10 -3 − x m 8.0 × 10 -4 − x m m m x
At equilibrium, ["FeSCN"^"2+"] = 1.7 × 10^"-4"color(white)(l) "mol/L" = x color(white)(l)"mol/L" [ FeSCN 2+ ] = 1.7 × 10 -4 l mol/L = x l mol/L
So x = 1.7 × 10^"-4" 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"
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+"
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