
The \[{{K}_{sp}}\] for $Fe{{(OH)}_{3}}=2.5\times {{10}^{-20}}$. How would you find the \[{{K}_{sp}}\] equation in terms of $F e^{+} 3$ and $O H^{-}$ concentrations?
Answer
545.7k+ views
Hint: To solve this question, we need to first find the chemical reaction. After finding the chemical reaction, we will then move ahead for solving the \[{{K}_{sp}}\] from the formula given below.
Formula Used:
We will use the following formula to find the required solution:
$K_{s p}=\left[A^{+}\right]^{a}\left[B^{-}\right]^{b}$
$K_{s p}$ = solubility product constant
$A^{+}=$ cation in an aqueous solution
$B^{-}=$ anion in an aqueous solution
$a, b=$ relative concentrations of $a, b=$ and $a, b=$
Complete step-by-step answer:We can represent the solubility of $Fe{{(OH)}_{3}}$ as:
$F e(O H)_{3}(s) \rightleftharpoons F e^{3+}+3 H O^{-}$
So, using the above formula, we have
$K_{s p}=\left[F e^{3+}\right]\left[H O^{-}\right]^{3}$
And if we let $S$ be the solubility of lead iodate
Then,
${{K}_{sp}}=(S){{(3S)}^{3}}=27{{S}^{4}}$
According to the question, we have been provided with the value of \[{{K}_{sp}}=2.5\times {{10}^{-20}}\]
Now, we will substitute this value in the above formula to get the result
\[27{{S}^{4}}=2.5\times {{10}^{-20}}\]
Upon solving, we get
\[\therefore S={}^{4}\sqrt{\dfrac{2.5\times {{10}^{-20}}}{27}}\]
Additional Information:
The product of solubility is a kind of constant of equilibrium and its value depends on temperature. Due to increased solubility, \[{{K}_{sp}}\]usually increases with a rise in temperature. Solubility depends on a number of parameters, the most important of which are salt lattice enthalpy and ion solvation enthalpy in the solution. Some significant factors that have an impact on the continuum of the solubility product are:
Common-ion impact (the presence of a common ion lowers the value of \[{{K}_{sp}}\]).
Diverse-ion impact (if the ions of the solutes are uncommon, the value of \[{{K}_{sp}}\]will be high).
The ion-pairs presence.
Note:Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution. The solubility in water of ionic compounds (which dissociate to form cations and anions) varies significantly. Some compounds are highly soluble, while others are highly insoluble and may even absorb moisture from the atmosphere.
Formula Used:
We will use the following formula to find the required solution:
$K_{s p}=\left[A^{+}\right]^{a}\left[B^{-}\right]^{b}$
$K_{s p}$ = solubility product constant
$A^{+}=$ cation in an aqueous solution
$B^{-}=$ anion in an aqueous solution
$a, b=$ relative concentrations of $a, b=$ and $a, b=$
Complete step-by-step answer:We can represent the solubility of $Fe{{(OH)}_{3}}$ as:
$F e(O H)_{3}(s) \rightleftharpoons F e^{3+}+3 H O^{-}$
So, using the above formula, we have
$K_{s p}=\left[F e^{3+}\right]\left[H O^{-}\right]^{3}$
And if we let $S$ be the solubility of lead iodate
Then,
${{K}_{sp}}=(S){{(3S)}^{3}}=27{{S}^{4}}$
According to the question, we have been provided with the value of \[{{K}_{sp}}=2.5\times {{10}^{-20}}\]
Now, we will substitute this value in the above formula to get the result
\[27{{S}^{4}}=2.5\times {{10}^{-20}}\]
Upon solving, we get
\[\therefore S={}^{4}\sqrt{\dfrac{2.5\times {{10}^{-20}}}{27}}\]
Additional Information:
The product of solubility is a kind of constant of equilibrium and its value depends on temperature. Due to increased solubility, \[{{K}_{sp}}\]usually increases with a rise in temperature. Solubility depends on a number of parameters, the most important of which are salt lattice enthalpy and ion solvation enthalpy in the solution. Some significant factors that have an impact on the continuum of the solubility product are:
Common-ion impact (the presence of a common ion lowers the value of \[{{K}_{sp}}\]).
Diverse-ion impact (if the ions of the solutes are uncommon, the value of \[{{K}_{sp}}\]will be high).
The ion-pairs presence.
Note:Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution. The solubility in water of ionic compounds (which dissociate to form cations and anions) varies significantly. Some compounds are highly soluble, while others are highly insoluble and may even absorb moisture from the atmosphere.
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