Write the overall net-ionic reaction that occurs when iron (ii) hydroxide (Fe(OH)2, Ksp = 7.9 x 10^-15) is dissolved in a 0.500 M sodium cyanide solution forming a complex ion. (Fe(CN)6^4-, Kf = 7.7 x 10^36)
Determine the value of the equilibrium constant for the overall net-ionic reaction.
Which chemical is the ligand in the complex ion?
Fe(OH)2 ==> Fe^2+ + 2OH^- Ksp = ?
Fe^2+ + 6CN^- ==> [Fe(CN6]^4 Kf = ?
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Fe(OH)2+ 6CN^- => Fe^2+ +2OH^- +[Fe(CN)6]^4-
CN^- is the ligand
Keq for the rxn as shown is Kf*Ksp = ?
do I have to find Ksp using the .5M?
No. K is K. It changes ONLY with temperature. Ksp is given Kf is given. If you wonder how I know Keq is Kf*Ksp, here is what you do.
Write the Ksp expression and multiply that by the Kf
To find the net-ionic reaction and equilibrium constant, we can start by writing the balanced equation for the dissolution of iron (II) hydroxide (Fe(OH)2) in sodium cyanide solution:
Fe(OH)2 + 6 NaCN → Na4[Fe(CN)6] + 2 NaOH
Now, let's write the net-ionic equation. This equation only includes the species that participate in the reaction:
Fe(OH)2 + 6 CN- → [Fe(CN)6]4- + 2 OH-
In this net-ionic equation, the sodium ions (Na+) and hydroxide ions (OH-) are spectator ions because they appear unchanged on both sides of the equation.
Now, let's determine the value of the equilibrium constant for the overall net-ionic reaction using the given formation constant (Kf) for the complex ion ([Fe(CN)6]4-):
Kf = [Fe(CN)6]4- / (CN-)6
Since the stoichiometric coefficient of CN- in the net ionic equation is 6, the value of the equilibrium constant (Kc) for the overall reaction is equal to the formation constant (Kf):
Kc = Kf = 7.7 x 10^36
Now, to identify the ligand in the complex ion [Fe(CN)6]4-, it is the group of atoms that coordinates with the central metal ion (Fe2+) to form the complex. In this case, the ligand is the cyanide ion (CN-), which coordinates with the iron (II) ion to form the complex ion [Fe(CN)6]4-.