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nignag [31]
1 year ago
8

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 ____________
Chemistry
1 answer:
Sindrei [870]1 year ago
7 0

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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