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Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility. the molar solubility is ___________ and the solubility is ____________

Sagot :

The molar solubility is 7.4×[tex]10^{-3}[/tex] M and the solubility is  7.4×[tex]10^{-3}[/tex] g/L .

Calculation ,

The dissociation of silver bromide is given as ,

[tex]AgBr[/tex] → [tex]Ag ^{+}[/tex] + [tex]Br^{-}[/tex]

S  

 -          S        S        

Ksp =  [[tex]Ag ^{+}[/tex] ] [ [tex]Br^{-}[/tex] ]  =  [S] [ S ] = [tex]S^{2}[/tex]

S = √ Ksp = √ 5. 5×[tex]10^{-5}[/tex] = 7.4×[tex]10^{-3}[/tex]

The solubility =7.4×[tex]10^{-3}[/tex] g/L

The molar solubility is the solubility of one mole of the substance.

Since ,  one mole of [tex]AgBr[/tex] is dissociates and form one mole of each  [tex]Ag ^{+}[/tex] and [tex]Br^{-}[/tex] ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4×[tex]10^{-3}[/tex] mol/L = 7.4×[tex]10^{-3}[/tex] M

To learn more about molar solubility ,

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