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This is one of our many quizzes on Solubility equilibrium. Click on this to access more such competitive quizzes on Physical Chemistry.

Solubility equilibrium Quiz 2

chemistry

 Solubility equilibrium is also known as dynamic equilibrium, can be seen when sparingly soluble salt is dissolved in water. Which of these is a sparingly soluble salt?

 The salt that will not dissociate completely is a sparingly soluble salt.

NaCl, KCl, and $MgC{{l}_{2}}$ are salts that are completely dissociated into ions when dissolved in water, but AgCl partially dissociates ions, therefore, it is a sparingly soluble salt.

 What is the solubility of $Ca{{F}_{2}}$ calcium fluoride, given that it is a sparingly soluble salt with a  ${{K}_{sp}}$ of $5.3\text{ x 1}{{\text{0}}^{-9}}$? 

 The given compound is $Ca{{F}_{2}}$, the reaction will be:$\\ \Rightarrow Ca{{F}_{2}}\rightleftharpoons C{{a}^{2+}}+2{{F}^{-}}\\ \Rightarrow $ The formula of solubility product for the given reaction will be:$ \\ \Rightarrow{{K}_{sp}}=[C{{a}^{2+}}]{{[2{{F}^{-}}]}^{2}}$

The given compound is $Ca{{F}_{2}}$, the reaction will be:<br>$\\  \Rightarrow Ca{{F}_{2}}\rightleftharpoons C{{a}^{2+}}+2{{F}^{-}}\\ \Rightarrow  \text{The formula of solubility product for the given reaction will be:}\\ \Rightarrow{{K}_{sp}}=[C{{a}^{2+}}]{{[2{{F}^{-}}]}^{2}}\\ \Rightarrow{{K}_{sp}}=(s){{(2s)}^{2}}\\ \Rightarrow{{K}_{sp}}=4{{s}^{3}}\\ \Rightarrow5.3\text{ x 1}{{\text{0}}^{-9}}=4{{s}^{3}}\\ \Rightarrow\dfrac{5.3\text{ x 1}{{\text{0}}^{-9}}}{4}={{s}^{3}} \\ \Rightarrow s=1.09\text{ x 1}{{\text{0}}^{-3}}$  

 Here are four salts with their respective solubility products, with the highest solubility being:<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4bbd6d5f-8794-4917-96ca-5269d8afb63c4622624196278067709.png"/>

  The dissociation increases with the value of the solubility product.

 From the given salts, the solubility product of $SrS{{O}_{4}}$ is the highest. So the solubility of  $SrS{{O}_{4}}$ will is the highest.

 For a compound, the formula of solubility product is given as:<br>${{K}_{sp}}=(s)\text{ x (3}s{{)}^{3}}$ <br>How many ions are present in this compound?

 We can calculate the ions by counting the number of s in the formula.

 There will be four ions as the given formula is ${{K}_{sp}}=(s)\text{ x (3}s{{)}^{3}}$. So, there are 1 cation and 3 anions.

 When the analysis was done, a saturated solution of sparingly soluble lead sulfate contains 10.80 grams /liter of the salt at room temperature. What will be ${{K}_{sp}}$ of the salt?

 First we have to calculate the solubility.$s=\dfrac{10.80}{Mass\text{ }of\text{ }PbS{{O}_{4}}}$ And then,${{K}_{sp}}=[P{{b}^{2+}}][SO_{4}^{2-}]$

 First we have to calculate the solubility.<br>$\\ s=\dfrac{10.80}{Mass\text{ }of\text{ }PbS{{O}_{4}}}\\ s=\dfrac{10.80}{303}=3.56\text{ x 1}{{\text{0}}^{-2}}\\ \text{The solubility product will be:}\\ PbS{{O}_{4}}\rightleftharpoons P{{b}^{2+}}+SO_{4}^{2-}]\\ {{K}_{sp}}=[P{{b}^{2+}}][SO_{4}^{2-}]\\ {{K}_{sp}}=(s)(s)=(3.56\text{ x 1}{{\text{0}}^{-2}})(3.56\text{ x 1}{{\text{0}}^{-2}})=1.27\text{ x 1}{{\text{0}}^{-3}}$ 