0.133 mg of silver bromide AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr?

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Explanation: 

The solubility equilibrium expression is :

AgBr (s) <-----> Ag+ (aq) + Br - (aq)

Solubility-product constant expression is :

Ksp= [Ag+][Br-]

The solubility is given as 0.133 mg/1.00 L, but Ksp uses molarity , so convert mg to g and then grams to moles by using the moalr mass  of the AgBr ( MM of AgBr is 187.72 )

0.133 mg/1.00 L x 1 mol / 187.72 = 7.083 x 10-7  M

So, as AgBr dissociates into 1 mol of each Ag+ and Br- ions the concentration of both ions would be 7.083 x 10-7  M.

Ksp= (7.083 × 10-7)2

Ksp= 5.02 × 10-13

 

 
Answer: 

Ksp= 5.02 × 10-13