Question

The concentration of $A{{g}^{+}}$ in a saturated solution of $A{{g}_{2}}C{{r}_{{}}}{{O}_{4}}$ at 293 K is given by $1.5\times {{10}^{-4}}mol\text{ }{{\text{l}}^{\text{-1}}}$. The solubility product of $A{{g}_{2}}C{{r}_{{}}}{{O}_{4}}$ at the same temperature is given by:
$\begin{aligned} & a)1.687\times {{10}^{-12}} \\\ & b)1.75\times {{10}^{-10}} \\\ & c)3.0\times {{10}^{-8}} \\\ & d)4.5\times {{10}^{-10}} \\\ \end{aligned}$

Answer 1

Hint: Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as ${{K}_{sp}}$.

Step-by-Step Solution:
Let us first define the term solubility product to help facilitate better understanding before we move onto the solution of this particular question.
The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol ${{K}_{sp}}$.
The solubility product is a kind of equilibrium constant and its value depends on temperature. ${{K}_{sp}}$usually increases with an increase in temperature due to increased solubility.
The solubility of ionic compounds (which dissociate to form cations and anions) in water varies to a great deal. Some compounds are highly soluble and may even absorb moisture from the atmosphere whereas others are highly insoluble.
Having established what the solubility product of a reaction really is and its significance, let us now move onto this question and its step-by-step solution.
To start with, the equation for the ionization of Silver Chromate is given by:
$A{{g}_{2}}Cr{{O}_{4}}\leftrightharpoons 2A{{g}^{+}}+CrO_{4}^{2-}$
Therefore, we analyse that this ionization of 1 mole of $A{{g}_{2}}C{{r}_{{}}}{{O}_{4}}$ gives out 2 moles of $A{{g}^{+}}$ and 1 of $CrO_{4}^{2-}$.

& \because [A{{g}^{+}}]=1.5\times {{10}^{-4}} \\\ & \therefore [CrO_{4}^{2-}]=\dfrac{1.5\times {{10}^{-4}}}{2}=0.75\times {{10}^{-4}} \\\ \end{aligned}$$ Expression for the solubility product of $A{{g}_{2}}Cr{{O}_{4}}$ is: $$\begin{aligned} & {{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{2-}] \\\ & \Rightarrow {{K}_{sp}}={{(1.5\times {{10}^{-4}})}^{2}}\times (0.75\times {{10}^{-4}}) \\\ & {{K}_{sp}}=1.6875\times {{10}^{-12}} \\\ & \text{Hence,the solubility product of }A{{g}_{2}}Cr{{O}_{4}}\text{ }is\text{ }1.6875\times {{10}^{-12}} \\\ \end{aligned}$$ Thus, we can conclude that the answer to this question is a). Note: While the solubility product is a type of equilibrium constant, it isn’t theoretically the same and the difference between the two must be understood to help completely understand the theory behind this question. Also, be very careful of the stoichiometric quotients and their resultant powers in the equilibrium equation as to get confused and mess up the calculation due to this reason is very common in equilibrium-type questions.