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Question: How do you use the \({{K}_{sp}}\) value to calculate the molar solubility of the following compound ...

How do you use the Ksp{{K}_{sp}} value to calculate the molar solubility of the following compound in pure water? BaSO4BaS{{O}_{4}}
 Ksp=1.07×10−10~{{K}_{sp}}=1.07\times {{10}^{-10}}

Explanation

Solution

We know that the reaction given is reversible reaction and hence equilibrium can be attained at one process, so we have to find equilibrium constant. For a given degree of dissociation; by creating an initial and by change in equilibrium table which we also known as ICE table, and by using this equation we can easily formulate expressions for equilibrium constant. For that we need to find a variable multiplied by R to which will be equal to equilibrium constant and solve a simple equation add.

Complete answer:
Firstly here we need to set up an ICE table based on equilibrium reaction which describes a way in which barium sulfate (BaSO4)(BaS{{O}_{4}}) dissolves in aqueous solution.

Thus the salt is considered insoluble in water so that the solution will contain a small amount of dissolved ion, since around most of these compounds will always remain dissociated as solid.
So the barium sulfate will be dissolve in small quantity to produce:

BaSO4(aq)BaS{{O}_{4(aq)}} ⇌ Ba(aq)2++SO4(aq)2−Ba_{(aq)}^{2+}+SO_{4(aq)}^{2-}

Similarly, for each every mole of a barium sulfate that being dissolve in solution we get 11 mole of a barium cation Ba(aq)2+Ba_{(aq)}^{2+} and 11 mole of a sulfate anion SO4(aq)2−SO_{4(aq)}^{2-}

The molar solubility of salt that is number of mole of a barium sulfate thus by dissolving to release the ion per liter of the solution so that we can use ICE table to write:

| BaSO4(aq)BaS{{O}_{4(aq)}}| Ba(aq)2+Ba_{(aq)}^{2+}| SO4(aq)2−SO_{4(aq)}^{2-}
---|---|---|---
I| −-| 00| 00
C| −-| (+s)(+s)| (+s)(+s)
E| −-| ss| ss

By, definition solubility product constant Ksp{{K}_{sp}} for these solubility equilibrium would be equal to
Ksp=[Ba2+]⋅[SO42−]{{K}_{sp}}=[B{{a}^{2+}}]\cdot [SO_{4}^{2-}]
The expression of Ksp{{K}_{sp}} it use equilibrium concentration given by;
Ksp=s×s=s2{{K}_{sp}}=s\times s={{s}^{2}}
Now that we have value of Ksp{{K}_{sp}} that is  1.07×10−10~1.07\times {{10}^{-10}}
 s2=1.07×10−10~{{s}^{2}}=1.07\times {{10}^{-10}}
Taking square root on both the sides;
 s2=1.07×10−10~\sqrt{{{s}^{2}}}=\sqrt{1.07\times {{10}^{-10}}}
 s=1.07×10−10~s=\sqrt{1.07\times {{10}^{-10}}}
s=1.03×10−5s=1.03\times {{10}^{-5}} Here, s represents molar solubility.

This means that when we add barium sulfate into the water; we can only dissolve 1.03×10−51.03\times {{10}^{-5}} mole for each and every liter of the solution.

Note: Note that equilibrium constant of chemical reaction is value of it is reaction quotient at the chemical equilibrium state approach by the dynamic chemical system after sufficient time has elapsed from which it is composition hasn’t any measurable tendency toward further change.