Question

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 ____________

Answer 1

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

Calculation ,

The dissociation of silver bromide is given as ,

$AgBr$ → $Ag ^{+}$ + $Br^{-}$

S  

 -          S        S        

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

S = √ Ksp = √ 5. 5×$10^{-5}$ = 7.4×$10^{-3}$

The solubility =7.4×$10^{-3}$ g/L

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

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

Molar solubility = 7.4×$10^{-3}$ mol/L = 7.4×$10^{-3}$ M

To learn more about molar solubility ,

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