Question

At$\text{ }18{}^\circ \text{C }$ , the solubility of $\text{ CdS }$ in water is $\text{ 6}\text{.33 }\times \text{ 1}{{\text{0}}^{-\text{15}}}\text{ M }$. What is the concentration of $\text{ C}{{\text{d}}^{\text{2+}}}\text{ }$ an ion in a solution of $\text{ pH = 1}$ saturated with $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas, in which concentration of $\text{ }{{\text{H}}_{\text{2}}}\text{S = 0}\text{.1 M }$ ? The product of the first and second ionization constants of $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ is $\text{ 1}\text{.1 }\times \text{ 1}{{\text{0}}^{-22}}\text{ }$ at this temperature.
A) $\text{ 6}\text{.343}\times \text{1}{{\text{0}}^{-8}}\text{ M }$
B) $\text{ 4}\text{.368}\times \text{1}{{\text{0}}^{-8}}\text{ M }$
C) $\text{ 4}\text{.34368}\times \text{1}{{\text{0}}^{-9}}\text{ M }$
D) $\text{ 3}\text{.636}\times \text{1}{{\text{0}}^{-8}}\text{ M }$

Answer 1

In a saturated solution, there exists a dynamic equilibrium between the excess of the solute and the ions in the solution. For a salt-like $\text{ }{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\text{ }$ , the solubility product $\text{ }{{\text{K}}_{\text{sp}}}\text{ }$ is given as,

& \text{ }{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\text{ }\rightleftharpoons \text{ x}{{\text{A}}^{\text{+}}}\text{(aq) + y}{{\text{B}}^{-}}(\text{aq)} \\\ & \text{ }\therefore {{\text{K}}_{\text{sp}}}\text{ = }{{\left[ {{\text{A}}^{\text{+}}} \right]}^{\text{x}}}\text{ }{{\left[ {{\text{B}}^{-}} \right]}^{y}} \\\ \end{aligned}$$ First calculate the solubility product of the $\text{ CdS }$ in absence of saturated gas. Then determine the sulphide ion concentration from its dissociation constant. **Complete step by step answer:** We are providing with the following data: The solubility of cadmium sulphide $\text{ CdS }$ in water is, $\text{ S = 6}\text{.33 }\times \text{ 1}{{\text{0}}^{-\text{15}}}\text{ M }$ Hydrogen potential of the solution, when saturated with $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas, is, $\text{ pH = 1}$ Concentration of hydrogen sulphide $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas is, $\text{ }{{\text{C}}_{{{\text{H}}_{\text{2}}}\text{S}}}\text{ = 0}\text{.1 M }$ The product of the first and second ionisation constant of $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ is, $\text{ }{{\text{K}}_{\text{a1}}}\text{ }\times \text{ }{{\text{K}}_{\text{a2}}}\text{ = 1}\text{.1 }\times \text{ 1}{{\text{0}}^{-22}}\text{ }$ We have to find the concentration of $\text{ C}{{\text{d}}^{\text{2+}}}\text{ }$ ions in the solution. The solubility of an $\text{ CdS }$ in absence of hydrogen sulphide is given as, $\text{ CdS }\to \text{ C}{{\text{d}}^{\text{2+}}}\text{ + }{{\text{S}}^{\text{2}-}}\text{ }$ Thus, in the absence of the $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas, the concentration of salt that is $\text{ CdS }$ is equal to the concentration of $\text{ C}{{\text{d}}^{\text{2+}}}\text{ }$ ion and sulphide ion${{\text{S}}^{\text{2}-}}$. This can be written as, $$\text{ }\left[ \text{CdS} \right]\text{= }\left[ \text{C}{{\text{d}}^{\text{2+}}} \right]\text{= }\left[ {{\text{S}}^{\text{2-}}} \right]=\text{ 6}\text{.33 }\times \text{ 1}{{\text{0}}^{-15}}\text{M }$$ The solubility product is equal to the product of the concentration of ions in the solution. For $\text{ CdS }$, the solubility product is as follows, $$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }\left[ \text{C}{{\text{d}}^{\text{2+}}} \right]\left[ {{\text{S}}^{2-}} \right]=\text{ }{{\left( 6.33\times {{10}^{-15}} \right)}^{2}}\text{ = 4 }\times \text{1}{{\text{0}}^{-29}}\text{ }$$ With the knowledge of the solubility product of salt, we can determine whether the substance is precipitated or not. For precipitate, the ionic product would be less than the solubility product. Here, $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ is used as a precipitating agent. It forms a precipitate of$\text{ CdS }$. The hydrogen sulphide is a diprotic acid. The product of the ionisation constant is equal to its equilibrium constant. Given as follows, $\begin{aligned} & \text{ }{{\text{H}}_{\text{2}}}\text{S }\rightleftharpoons \text{ H}{{\text{S}}^{-}}\text{ + }{{\text{H}}^{\text{+}}}\text{ }{{\text{K}}_{\text{a1}}} \\\ & \text{ H}{{\text{S}}^{-}}\rightleftharpoons \text{ }{{\text{S}}^{2-}}+\text{ }{{\text{H}}^{\text{+}}}\text{ }{{\text{K}}_{\text{a2}}} \\\ \end{aligned}$ Thus, for hydrogen sulphide, $$\begin{aligned} & \text{ }{{\text{K}}_{\text{a1}}}\text{ }\times \text{ }{{\text{K}}_{\text{a2}}}\text{ = }\dfrac{{{\left[ {{\text{H}}^{\text{+}}} \right]}^{2}}\left[ {{\text{S}}^{2-}} \right]}{\left[ {{\text{H}}_{\text{2}}}\text{S} \right]}\text{ = 1}\text{.1 }\times \text{ 1}{{\text{0}}^{-22}}\text{ } \\\ & \text{ }\therefore \left[ {{\text{S}}^{2-}} \right]\text{ = }\dfrac{\text{1}\text{.1 }\times \text{ 1}{{\text{0}}^{-22}}\times \left[ {{\text{H}}_{\text{2}}}\text{S} \right]}{{{\left[ {{\text{H}}^{\text{+}}} \right]}^{2}}}\text{ = }\dfrac{\text{1}\text{.1 }\times \text{ 1}{{\text{0}}^{-22}}\times \text{ 0}\text{.1}}{{{\left( 0.1 \right)}^{2}}}\text{ = 1}\times \text{1}{{\text{0}}^{-21}}\text{ } \\\ \end{aligned}$$ Thus, using the solubility product formula the concentrations of cadmium ion when hydrogen sulphide gas is passed over would be equal to, $\begin{aligned} & \text{ }{{\text{K}}_{\text{sp}}}\text{ = }\left[ \text{C}{{\text{d}}^{\text{2+}}} \right]\left[ {{\text{S}}^{2-}} \right] \\\ & \Rightarrow \left[ \text{C}{{\text{d}}^{\text{2+}}} \right]=\text{ }\dfrac{{{\text{K}}_{\text{sp}}}}{\left[ {{\text{S}}^{2-}} \right]}\text{ = }\dfrac{4\times {{10}^{-29}}}{1.1\times {{10}^{-21}}}\text{ = 3}\text{.636}\times \text{1}{{\text{0}}^{-8}}\text{ M } \\\ \end{aligned}$ Therefore, the concentration of cadmium ion is, $\text{ }\left[ \text{C}{{\text{d}}^{\text{2+}}} \right]\text{= 3}\text{.636}\times \text{1}{{\text{0}}^{-8}}\text{ M }$ **Hence, (D) is the correct option.** **Note:** The hydrogen sulphide is a weak acid. It is dissociated into the hydrogen ion and sulphide ion. In an inorganic analysis, the dissociation of $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ is suppressed by the addition of $\text{ HCl }$ (common ion effect). Thus the concentration of sulphide ion decreases than the solubility product. Here we have observed the same. Because of these cations like cadmium, copper is easily precipitated as sulphides.