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**Hint:**In solution, we will see that if the ionic product will be greater than the solubility product, then ions will get precipitated. We will find equilibrium constant K. Equilibrium constant is denoted by the symbol $\left( {{k}_{c}} \right)$. It is found that the expression for equilibrium constant K is given by the equation:

$K=\dfrac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}$

**Complete Step by step solution:**

- $P{{b}^{2+}},Z{{n}^{2+}},N{{i}^{2+}}$ are equal and their concentration is 0.1 M. And if we will pass ${{H}_{2}}S$ gas through this solution, then ${{H}_{2}}S$ will become soluble in water and we will get:

\[{{H}_{2}}S\rightleftarrows 2H+{{S}^{2-}}\]

And for this reaction equilibrium constant will become:

$K={{K}_{1}}\times {{K}_{2}}$

We are being provided with values of ${{K}_{1}}{{H}_{2}}S={{10}^{-7}}$,${{K}_{2}}{{H}_{2}}S={{10}^{-14}}$

\[\begin{align}

& K={{10}^{-7}}\times {{10}^{-14}} \\

&\implies ={{10}^{-21}} \\

\end{align}\]

- The expression of equilibrium constant K can be written as:

\[K=\dfrac{{{\left[ {{H}^{+}} \right]}^{2}}\left[ {{S}^{2-}} \right]}{\left[ {{H}_{2}}S \right]}\]

We can see here that ${{K}_{1}}>>{{K}_{2}}$which means that concentration of ${{H}_{2}}S$:${{H}_{2}}S=\left[ {{H}^{+}} \right]={{10}^{-7}}M$

Now,

\[\begin{align}

& {{10}^{-21}}=\left[ {{H}^{+}} \right]\left[ {{S}^{2-}} \right] \\

& \Rightarrow \left[ {{S}^{2-}} \right]={{10}^{-14}}M \\

\end{align}\]

This is the concentration of ions present in the solution.

- As we know that for precipitation of any salt ionic product of salt ${{Q}_{sp}}$ should be greater than solubility product constant${{K}_{sp}}$. That is:

${{Q}_{sp}}\text{ }is\text{ }greater\text{ }than\text{ }{{K}_{sp}}$

- So, we will check the ionic product for PbS salt that is equal to the concentration of products of ions.

\[\begin{align}

& PbS=\left[ P{{b}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-14}} \\

& \implies ={{10}^{-15}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}PbS=3\times {{10}^{-29}}$, which means ionic product for PbS is > solubility product constant of PbS.

Hence, PbS precipitate will be formed.

- Similarly, we will discuss for ZnS:

\[\begin{align}

& ZbS=\left[ Z{{n}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-14}} \\

&\implies ={{10}^{-15}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}ZnS={{10}^{-25}}$, which means ionic product for ZnS is > solubility product constant of ZnS.

Hence, ZnS precipitate will be formed.

- Similarly, we will discuss for NiS:

\[\begin{align}

& NiS=\left[ N{{i}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-14}} \\

& \implies ={{10}^{-15}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}NiS=3\times {{10}^{-19}}$, which means ionic product for NiS is > solubility product constant of NiS.

Hence, NiS precipitate will be formed.

- Hence, we can say that assertion is true statement that is $P{{b}^{2+}},Z{{n}^{2+}},N{{i}^{2+}}$will get precipitated simultaneously.

- Now, let’s discuss about the reason:

If we will add 0.1M HCl in the solution, then as HCl is a strong acid then it will be completely ionised in the solution. And we will get 0.1 M HCl in the solution.

And we have seen from the equilibrium of ${{H}_{2}}S$ that $K={{10}^{-21}}$

Hence,

\[\begin{align}

& K={{10}^{-21}}=\dfrac{{{\left[ 0.1 \right]}^{2}}\left[ {{S}^{2-}} \right]}{\left[ 0.1 \right]} \\

& \left[ {{S}^{2-}} \right]={{10}^{-20}}M \\

\end{align}\]

Now, we will check which ions will be precipitated from the given ions.

- For, PbS ionic product ${{Q}_{sp}}$ will be:

\[\begin{align}

& PbS=\left[ P{{b}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-20}} \\

&\implies ={{10}^{-21}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}PbS=3\times {{10}^{-29}}$, which means ionic product for PbS is > solubility product constant of PbS.

Hence, $P{{b}^{2+}}$ precipitate will be formed.

- Similarly, we will discuss for NiS:

\[\begin{align}

& NiS=\left[ N{{i}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-20}} \\

&\implies ={{10}^{-21}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}NiS=3\times {{10}^{-19}}$, which means ionic product for NiS is

- Similarly, we will discuss for ZnS:

\[\begin{align}

& ZbS=\left[ Z{{n}^{2+}} \right]\left[ {{S}^{2-}} \right] \\

&\implies =0.1\times {{10}^{-20}} \\

& \implies={{10}^{-21}} \\

\end{align}\]

We are being provided with ${{K}_{sp}}ZnS={{10}^{-25}}$, which means ionic product for ZnS is > solubility product constant of ZnS.

Hence, $Z{{n}^{2+}}$ precipitate will be formed.

**- Therefore, we can conclude that the correct option is (a) that is both assertion and reason are correct and reason is the correct explanation for assertion.**

**Note:**- We should note here the main difference in Solubility product and ionic product. Solubility products are applicable only to saturated solutions, and at a constant temperature, its value is found to be constant for given electrolyte.

- Whereas, an ionic product is applicable to all solutions having any concentration, and its value also keeps changing as the concentration changes.

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