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.
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