Answer
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Hint: To solve this question, first we will write all three equations of reaction of $SC{N^ - }$ with $F{e^{3 + }}$ and then adding them we will get the equation of formation and then according the equilibrium constants will be multiplied with each other to get overall formation constant.
Complete step by step answer:
In this question, by formation constant we mean that constant which describes the formation of a complex ion from its central ion and attached ligands. It is represented as ${K_f}$.
Also, dissociation constant is that constant which describes the formation of central ions and attached ligands from its complex. It is represented as ${K_d}$. It is the inverse of formation constant.
Now, we will write the equations of formation when $SC{N^ - }$ reacts with $F{e^{3 + }}$and then again the formed product will react with $SC{N^ - }$ for the corresponding equilibrium constants. These equations are:
$
(1){\text{ }}F{e^{3 + }} + SC{N^ - } \to {[Fe(SCN)]^{2 + }}{\text{ }}{K_1} = 130 \\
(2){\text{ [}}Fe(SCN){]^{2 + }} + SC{N^ - } \to {[Fe{(SCN)_2}]^{ - 1}}{\text{ }}{K_2} = 16 \\
(3){\text{ }}{[Fe(SCN)]^{ - 1}} + SC{N^ - } \to [Fe{(SCN)_3}]{\text{ }}{K_3} = 1 \\
$
Now, we can write the overall equation by adding all above three equations:
$F{e^{3 + }} + 3SC{N^ - } \to Fe{(SCN)_3}$
As we get above equation by adding all equations hence, the overall formation constant is the product of formation constant of each above given equation that is :
$
{K_f} = {K_1}{K_2}{K_3} = 130 \times 16 \times 1 \\
{K_f} = 2080 \\
\\
$
Now, to find dissociation constant we will write dissociation equation that is :
$Fe{(SCN)_3} \to F{e^{3 + }} + 3SC{N^ - }$
As it is the inverse of overall formation equation hence, dissociation constant can be given as :
${K_d} = \dfrac{1}{{{K_f}}} = \dfrac{1}{{2080}}$
${K_d} = 4.8 \times {10^{ - 4}}$
Hence, it is the required answer.
Note:
A stability constant (formation constant, binding constant) is an equilibrium constant for the formation of a complex in solution. It is a measure of the strength of the interaction between the reagents that come together to form the complex.
Complete step by step answer:
In this question, by formation constant we mean that constant which describes the formation of a complex ion from its central ion and attached ligands. It is represented as ${K_f}$.
Also, dissociation constant is that constant which describes the formation of central ions and attached ligands from its complex. It is represented as ${K_d}$. It is the inverse of formation constant.
Now, we will write the equations of formation when $SC{N^ - }$ reacts with $F{e^{3 + }}$and then again the formed product will react with $SC{N^ - }$ for the corresponding equilibrium constants. These equations are:
$
(1){\text{ }}F{e^{3 + }} + SC{N^ - } \to {[Fe(SCN)]^{2 + }}{\text{ }}{K_1} = 130 \\
(2){\text{ [}}Fe(SCN){]^{2 + }} + SC{N^ - } \to {[Fe{(SCN)_2}]^{ - 1}}{\text{ }}{K_2} = 16 \\
(3){\text{ }}{[Fe(SCN)]^{ - 1}} + SC{N^ - } \to [Fe{(SCN)_3}]{\text{ }}{K_3} = 1 \\
$
Now, we can write the overall equation by adding all above three equations:
$F{e^{3 + }} + 3SC{N^ - } \to Fe{(SCN)_3}$
As we get above equation by adding all equations hence, the overall formation constant is the product of formation constant of each above given equation that is :
$
{K_f} = {K_1}{K_2}{K_3} = 130 \times 16 \times 1 \\
{K_f} = 2080 \\
\\
$
Now, to find dissociation constant we will write dissociation equation that is :
$Fe{(SCN)_3} \to F{e^{3 + }} + 3SC{N^ - }$
As it is the inverse of overall formation equation hence, dissociation constant can be given as :
${K_d} = \dfrac{1}{{{K_f}}} = \dfrac{1}{{2080}}$
${K_d} = 4.8 \times {10^{ - 4}}$
Hence, it is the required answer.
Note:
A stability constant (formation constant, binding constant) is an equilibrium constant for the formation of a complex in solution. It is a measure of the strength of the interaction between the reagents that come together to form the complex.
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