Answer
425.1k+ views
Hint: The solubility product is a constant and going to represent with \[{{K}_{sp}}\]. It is the equilibrium constant for a chemical which is dissolving in an aqueous solution. Solubility product signifies the level at which a solute dissolves in a solution. If a substance has more solubility, then the substance has higher \[{{K}_{sp}}\] value.
Complete step by step answer:
In the question it is given that the solubility product (\[{{K}_{sp}}\]) of lead iodide is \[1.4\times {{10}^{-8}}\].
The molecular formula of lead iodide is\[Pb{{I}_{2}}\].
\[Pb{{I}_{2}}\] ionizes in an aqueous solution as follows.
\[Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}}\]
Assume the solubility of \[Pb{{I}_{2}}\] is S. Then
\[\begin{align}
& Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}} \\
& S \quad \quad \quad S \quad \quad \quad 2S \\
\end{align}\]
We have to calculate the solubility of lead iodide in 0.1 M KI (potassium iodide) solution.
We know that potassium iodide is a strong electrolyte and it will ionize completely in an aqueous solution and releases iodide ions (\[{{I}^{-}}\]=0.1 M).
Therefore the concentration of lead and iodide ions in an aqueous solution is as follows.
\[\begin{align}
& [P{{b}^{2+}}]=S\text{ and } \\
& [{{I}^{-}}]=(2S+0.1)M \\
\end{align}\](Here S = solubility of respective ions)
Now we have to substitute all these values in solubility product formula
\[\begin{align}
& {{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}} \\
& 1.4\times {{10}^{-8}}=\text{ }S\text{ }\times \text{ }{{(2S+0.1)}^{2}} \\
& S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}} \\
\end{align}\]
Therefore the molar solubility of lead iodide in 0.1 M KI solution is \[S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}}\].
Note:
Iodide is going to release from lead iodide and potassium iodide into the aqueous solution. So, whenever we are going to take the concentration of iodide in the solution we have to include the concentration of iodide from both lead iodide and potassium iodide.
That’s why\[[{{I}^{-}}]=(2S+0.1)M\].
2S = from lead iodide
And 0.1 = form potassium iodide.
Complete step by step answer:
In the question it is given that the solubility product (\[{{K}_{sp}}\]) of lead iodide is \[1.4\times {{10}^{-8}}\].
The molecular formula of lead iodide is\[Pb{{I}_{2}}\].
\[Pb{{I}_{2}}\] ionizes in an aqueous solution as follows.
\[Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}}\]
Assume the solubility of \[Pb{{I}_{2}}\] is S. Then
\[\begin{align}
& Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}} \\
& S \quad \quad \quad S \quad \quad \quad 2S \\
\end{align}\]
We have to calculate the solubility of lead iodide in 0.1 M KI (potassium iodide) solution.
We know that potassium iodide is a strong electrolyte and it will ionize completely in an aqueous solution and releases iodide ions (\[{{I}^{-}}\]=0.1 M).
Therefore the concentration of lead and iodide ions in an aqueous solution is as follows.
\[\begin{align}
& [P{{b}^{2+}}]=S\text{ and } \\
& [{{I}^{-}}]=(2S+0.1)M \\
\end{align}\](Here S = solubility of respective ions)
Now we have to substitute all these values in solubility product formula
\[\begin{align}
& {{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}} \\
& 1.4\times {{10}^{-8}}=\text{ }S\text{ }\times \text{ }{{(2S+0.1)}^{2}} \\
& S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}} \\
\end{align}\]
Therefore the molar solubility of lead iodide in 0.1 M KI solution is \[S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}}\].
Note:
Iodide is going to release from lead iodide and potassium iodide into the aqueous solution. So, whenever we are going to take the concentration of iodide in the solution we have to include the concentration of iodide from both lead iodide and potassium iodide.
That’s why\[[{{I}^{-}}]=(2S+0.1)M\].
2S = from lead iodide
And 0.1 = form potassium iodide.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)