Question

0.1 milli mole of $CdS{{O}_{4}}$ are present in a 10 mL acid solution of 0.08 N HCl. Now ${{H}_{2}}S$ is passed to precipitate all the $C{{d}^{2+}}$ ions. The pH of solution after filtering off precipitate, boiling of ${{H}_{2}}S$ and making the solution 100 mL by adding ${{H}_{2}}O$ will be:
(A) 2
(B) 4
(C) 8
(D) None of these

Answer 1

Hint: pH of a solution is calculated from the concentration of ${{H}^{+}}$ ions in the solution and is given as $pH=-\log \left[ {{H}^{+}} \right]$.
${{H}_{2}}S$ precipitates all the $C{{d}^{2+}}$ ions in the solution due to $CdS{{O}_{4}}$ by the following reaction
     \[CdS{{O}_{4}}+{{H}_{2}}S\to CdS\downarrow +{{H}_{2}}S{{O}_{4}}\]


Complete step by step answer:
We have been given that 0.1$\times {{10}^{-3}}$ mole of cadmium sulphate ($CdS{{O}_{4}}$) is present in 10 mL solution of 0.08N hydrochloric acid (HCl).
Normality of a solution is given
     \[\text{Normality (N) = }\dfrac{\text{number of mole equivalent (n)}}{\text{volume of solution (in mL)}}\times 1000\]

Let us now calculate the number moles of HCl present in 10 mL of 0.08N solution.
Given, normality of HCl solution, N = 0.08 N.
Volume of HCl solution, V = 10 mL
The number of moles of HCl, n =\[\dfrac{N\times V}{1000}\]
 Substituting the values of normality and volume of HCl solution in the above equation, we get
     \[n=\dfrac{0.08\times 10}{1000}=0.8\times {{10}^{-3}}\]

Now, hydrogen sulphide (${{H}_{2}}S$) is made to pass through the solution. ${{H}_{2}}S$ will precipitate out the $C{{d}^{2+}}$ ions in the form of cadmium sulphide (CdS). HCl is acting as a catalyst in the reaction. The chemical equation for precipitation of $C{{d}^{2+}}$ ions by ${{H}_{2}}S$ is given below:
     \[CdS{{O}_{4}}+{{H}_{2}}S+HCl\to CdS\downarrow +{{H}_{2}}S{{O}_{4}}+HCl\]

 Since HCl is acting as a catalyst in the reaction, therefore, its concentration remains the same.
\[\begin{matrix}
   {} & CdS{{O}_{4}} & + & {{H}_{2}}S & \to & CdS & + & {{H}_{2}}S{{O}_{4}} \\
   Initial\,number\,of\,moles & 0.1\times {{10}^{-3}} & {} & 0.1\times {{10}^{-3}} & {} & 0 & {} & 0 \\
   Final\,number\,of\,moles & 0 & {} & 0 & {} & 0.1\times {{10}^{-3}} & {} & 0.1\times {{10}^{-3}} \\
\end{matrix}\]

After ${{H}_{2}}S$ has completely precipitated $C{{d}^{2+}}$ ions, the solution is filtered to obtain the precipitates. The remaining ${{H}_{2}}S$ in the solution is boiled and water is added to make the volume of the solution up to 100 mL.

Now the solution contains only HCl and ${{H}_{2}}S{{O}_{4}}$.
Number of moles of HCl = 0.8$\times {{10}^{-3}}$ mole
Number of moles of ${{H}_{2}}S{{O}_{4}}$ = 0.1$\times {{10}^{-3}}$ mole.

Now, we know that HCl being a strong acid dissociates completely in water into ${{H}^{+}}$ and $C{{l}^{-}}$ ions
     \[HCl(aq)\to {{H}^{+}}(aq)+C{{l}^{-}}(aq)\]

Since one mole of HCl gives one mole of ${{H}^{+}}$ and $C{{l}^{-}}$ ions. Thus, the number of moles of ${{H}^{+}}$ will be = 0.1$\times {{10}^{-3}}$ mole.
Similarly, ${{H}_{2}}S{{O}_{4}}$ dissociates completely into ${{H}^{+}}$ and $SO_{4}^{2-}$ as
     \[{{H}_{2}}S{{O}_{4}}(aq)\to 2{{H}^{+}}(aq)+SO_{4}^{2-}(aq)\]

Now, one mole of ${{H}_{2}}S{{O}_{4}}$ gives two moles of ${{H}^{+}}$ so, moles of ${{H}^{+}}$ ions = 2$\times $ 0.1$\times {{10}^{-3}}$ mole.
Therefore, the number of number of ${{H}^{+}}$ ions is equal to the sum of number of moles of ${{H}^{+}}$ ions from HCl and ${{H}_{2}}S{{O}_{4}}$, i.e.

Total number of moles of ${{H}^{+}}$ = (0.1$\times {{10}^{-3}}$ + 2$\times $ 0.1$\times {{10}^{-3}}$) mole = 1$\times {{10}^{-3}}$ mole = 1 millimole.

To find the pH of the solution, we need to find the concentration of ${{H}^{+}}$ in moles per liter which is equal to the number of moles of ${{H}^{+}}$ ions divided by volume of the solution.
We have given that the volume of the solution is made up to 100 ml. Therefore, the concentration of ${{H}^{+}}$ ions will be equal to
     \[\left[ {{H}^{+}} \right]=\dfrac{1\times {{10}^{-3}}mol}{100\times {{10}^{-3}}L}={{10}^{-2}}mol{{L}^{-1}}\]

Thus, the pH of the solution will be calculated as follows
     $\begin{align}
  & pH=-\log \left[ {{H}^{+}} \right] \\
 & pH=-\log \left[ {{10}^{-2}} \right] \\
\end{align}$
Applying $\log {{m}^{n}}=n\log m$, we get
     $pH=-\log \left[ {{10}^{-2}} \right]=2$
Therefore, the pH of the solution is 2.
So, the correct answer is “Option A”.

Note: It is to be noted that HCl is not consumed during the reaction. We are likely to get confused with the fact that HCl is not reacting in the reaction. So remember that concentration of a catalyst does not change, it is recovered at the completion of the reaction in the same amount. Carefully solve the question step by step to avoid any confusion or error.