Question

Which of the following is most soluble in water?
(A)- $B{{i}_{2}}{{S}_{3}}({{K}_{sp}}=1\times {{10}^{-70}}{{M}^{5}})$
(B)- $MnS({{K}_{sp}}=7\times {{10}^{-16}}{{M}^{2}})$
(C)- $CuS({{K}_{sp}}=8\times {{10}^{-37}}{{M}^{2}})$
(D)- $A{{g}_{2}}S({{K}_{sp}}=6\times {{10}^{-51}}{{M}^{3}})$

Answer 1

Hint: The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the solubility product constant, then the compound is more soluble.

Complete step by step solution:
We know that formula for solubility.
Let’s consider “s” as solubility and ${{K}_{sp}}$as a solubility product constant.
The solubility products have different units. So, we cannot compare them directly. Therefore, we will calculate the solubility in each case.
Then whichever solubility will be maximum, it will be our answer.

Now, for $B{{i}_{2}}{{S}_{3}}$, we will assume ‘s’ be the solubility of each of the ions formed. $B{{i}_{2}}{{S}_{3}}$ will dissociate into ions according to the below chemical equation:
$B{{i}_{2}}{{S}_{3}}\to 2B{{i}^{+3}}+3{{S}^{-2}}$
Therefore, the solubility product for $B{{i}_{2}}{{S}_{3}}$ will be:
${{K}_{sp}}={{(2s)}^{2}}{{(3s)}^{3}}=108{{s}^{5}}$
Hence, ‘s’ will be calculated by taking the root of the ${{K}_{sp}}$ and dividing it by 108

{{s}^{5}}=\dfrac{{{K}_{sp}}}{108} \\\ s=\sqrt[5]{\dfrac{{{K}_{sp}}}{108}} \\\ \begin{aligned} & s=\dfrac{\sqrt[5]{{{K}_{sp}}}}{108} \\\ & {{K}_{sp}}=1\times {{10}^{-70}}{{M}^{5}} \\\ \end{aligned} \\\ s=\dfrac{\sqrt[5]{1\times {{10}^{-70}}}}{108} \\\ s=2\times {{10}^{-14}}M \\\ \end{array}$$ Now, we will be calculating for option (B) $$MnS\to M{{n}^{+2}}+{{S}^{-2}}$$ Therefore, $$\begin{aligned} & {{K}_{sp}}=s\times s={{s}^{2}} \\\ & s=\sqrt{{{K}_{sp}}} \\\ \end{aligned}$$ $$\begin{array}{*{35}{l}} MnS: \\\ s=\sqrt{7\times {{10}^{-16}}}=2.6\times {{10}^{-8}}M \\\ \end{array}$$ Now, we will be calculating for option (C) $$CuS\to C{{u}^{+2}}+{{S}^{-2}}$$ Therefore, $$\begin{aligned} & {{K}_{sp}}=\,s\times s={{s}^{2}} \\\ & s=\sqrt{{{K}_{sp}}} \\\ \end{aligned}$$ $\begin{array}{*{35}{l}} Cus: \\\ s=\sqrt{{{K}_{sp}}}=\sqrt{8\times {{10}^{-37}}}=8.95\times {{10}^{-19}}M \\\ \end{array}$ Now, we will be calculating for option (D) $$A{{g}_{2}}S\to 2A{{g}^{+1}}+{{S}^{-2}}$$ Therefore, $$\begin{aligned} & {{K}_{sp}}=\,{{(2s)}^{2}}(s)=4{{s}^{3}} \\\ & s\,=\,\sqrt[3]{\dfrac{{{K}_{sp}}}{4}} \\\ \end{aligned}$$ $$\begin{array}{*{35}{l}} A{{g}_{2}}{{S}_{{}}}: \\\ {{K}_{sp}}=6\times {{10}^{-51}}{{M}^{3}} \\\ s=\dfrac{\sqrt[3]{{{K}_{sp}}}}{4} \\\ s=\dfrac{\sqrt[3]{6\times {{10}^{-51}}}}{4} \\\ s=4.54\times {{10}^{-18}}M \\\ \end{array}$$ According to the above calculations, we can conclude that the option (B) is the correct answer. Additional Information: Lets see the importance of solubility and solubility product constant and relation between solubility and solubility. Solubility product constant is the equilibrium between a solid and its respective ions in a solution. The value of the solubility product constant identifies the degree of which the compound can dissociate in water. Solubility product constant is used to describe the saturated solution of ionic compounds. (A saturated solution is when there is a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound). The relation between solubility and the solubility product constants is that one can be used to find the other. In other words, there is a relationship between the solution's molarity and the solubility of the ions because solubility product constants is literally the product of the solubility of each ion in moles per liter. Note: The relation between solubility and solubility product constant is quite important when describing the solubility of slightly ionic compounds. Mentioning ionic compounds because most ionic compounds are soluble in water or other solvents as to where molecular compounds are not soluble.