The balanced equation is
#"M"^"2+" + "6CN"^"-" ⇌ "M(CN)"_6^"4+"; K_text(f) = 2.50 × 10^17#
We can set up an ICE table.
#color(white)(mmmmmmml)"M"^"2+" + color(white)(ll)"6CN"^"-" ⇌ "M(CN)"_6^"4+"#
#"I/mol·L"^"-1": color(white)(mll)0.150 color(white)(mm)1.220color(white)(mmmll)0#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmll)"-6"xcolor(white)(mmml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.150-"xcolor(white)(m)"1.220-6"xcolor(white)(mml)x#
#K_text(f) = (["M(CN)"_6^"4+"])/(["M"^"2+"]["CN"^"-"]^6) = x/(("0.150-"x)(1.220-6"x)^6) = 2.50 × 10^17#
This result looks rather unpromising. We must solve a seventh-order polynomial.
However, the value of #K_text(f)# is so large that the reaction goes substantially to completion.
Thus, 0.150 mol of #"M"^"2+"# will react with 0.900 mol of #"CN"^"-"# and form 0.150 mol of #"M(CN)"_6^"4+"# (with 0.300 mol of #"CN"^"-"# unused).
We can approach the equilibrium from the other end. Let's assume that the reaction goes to completion and write the equilibrium in reverse.
#"M(CN)"_6^"4+" ⇌"M"^"2+" + "6CN"^"-"#
Then,
#K = 1/K_text(f) = 1/(2.50 × 10^17) = 4.00× 10^"-18"#
Our new ice table becomes
#color(white)(mmmmmm)"M(CN)"_6^"4+" ⇌ "M"^"2+" + "6CN"^"-"#
#"I/mol·L"^"-1": color(white)(mll)0.150 color(white)(llmmm)0color(white)(mml)0.300#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmm)"+"xcolor(white)(mm)"+6"x#
#"E/mol·L"^"-1": color(white)(m)"0.150-"xcolor(white)(mmm)xcolor(white)(ml)"0.300+6"x#
#K = (["M"^"2+"]["CN"^"-"]^6)/(["M(CN)"_6^"4+"]) = (x("0.300+6"x)^6)/("0.150-"x)= 4.00 × 10^"-18"#
Check for negligibility
#0.300/(4.00 × 10^"-18") = 7. 5 × 10^16 > 400#. ∴ #6x ≪ 0.300# and #x ≪ 0.150#.
Then,
#(x(0.300)^6)/0.150 = 4.00 × 10^"-18"#
#7.29 × 10^"-4"x = 0.150 × 4.00 × 10^"-18" = 6.00 × 10^"-19"#
#x = (6.00 × 10^"-19")/(7.29 × 10^"-4") = 8.23 × 10^"-16"#
#["M"^"2+"] = 8.23 × 10^"-16"color(white)(l)"mol/L"#
Check:
#K = ((8.23 × 10^"-16")(0.300 + 6 × 8.23 × 10^"-16")^6)/("0.150 -8.23 × 10"^"-16") = ((8.23 × 10^"-16")(0.300)^6)/0.150 = 4.00 × 10^"-18"#
It checks!
