Question

The ratio of amount of \[{H_2}S\] needed to precipitate the metal ion from \[20ml\] each of \[1M{\text{ Cd(N}}{{\text{O}}_{\text{3}}}{\text{)}}{\,_2}\] and \[0.5{\text{M CuS}}{{\text{O}}_4}\] is ?
A.\[1:1\]
B.\[2:1\]
C.\[1:2\]
D.Indefinite

Answer 1

Hint: Molar amount or molar concentration is basically the amount of substance or solute present in the volume of the solution. We can also simply term it as the Molarity. In this case, where we have to find the ratio of the molar concentrations between two equations, we will just compare them by finding the molar amount of $H_2S$ produced in both the equations separately.

Complete answer:
To simplify this question, first we need to know the proper balanced equation between the cadmium nitrate, \[Cd{(N{O_3})_2}\] and hydrogen sulphide, \[{H_2}S\] and in the second equation between copper sulphate, \[CuS{O_4}\] and hydrogen sulphide.
So let’s begin by writing the equations –
\[Cd{(N{O_3})_2} + {H_2}S \to CdS + 2HN{O_3}......(i)\]
So, by this we get that, \[1{\text{ mol}}\] of \[Cd{(N{O_3})_2}\] reacts with \[1{\text{ mol}}\]of \[{H_2}S\]
Since we need to find for \[20ml\] of 1M \[Cd{(N{O_3})_2}\] so,
Our formula will be,
 \[\begin{gathered}
  20ml{\text{ Cd(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{ = }}\dfrac{{20}}{{1000}}l{\text{ }} \times {\text{1mol}}{{\text{L}}^{{\text{ - 1}}}} \\
   = 0.02mol \\
\end{gathered} \]
Hence the required \[{H_2}S\] for \[20ml{\text{ Cd(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{ }}\]is \[0.02{\text{ }}mol\]
Now, we will find the amount of \[{H_2}S\] for \[CuS{O_4}\]
So, first we will write the balanced chemical equation between copper sulphate and hydrogen sulphide
\[CuS{O_4} + {H_2}S \to CuS + {H_2}S{O_4}......(ii)\]
This states that \[1{\text{ mol}}\] of \[CuS{O_4}\] reacts with \[1mol\] of \[{H_2}S\]
Now as we did above, we will find the amount of $H_2S$ for 20 ml of 0.5M \[CuS{O_4}\]
Hence our formula will be
\[\begin{gathered}
  20ml{\text{ of 0}}{\text{.5M CuS}}{{\text{O}}_{\text{4}}}{\text{ = }}\dfrac{{20}}{{1000}} \times 0.5{\text{ mol}}{{\text{L}}^{{\text{ - 1}}}} \\
   = 0.01{\text{ mol }} \\
\end{gathered} \]
So, the amount of $H_2S$ required by \[CuS{O_4}\] to precipitate metal is \[0.01{\text{ }}mol\].
Now we will move to our final step
We will compare both the molar amount of h2s produced by reacting with \[Cd{(N{O_3})_2}\] and \[CuS{O_4}\] Hence, the ratio between \[{H_2}S\] is -
\[\dfrac{{0.02}}{{0.01}} = 2:1\]
So, our desired ratio is \[2:1\], hence we can say option B is correct.

Note:
While finding the molar concentrations make sure to always have a balanced equation, this helps us to know about the correct amount of substance required to react with a particular chemical species.