Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
Homework Answers
The concept used in this problem is based on the solubility product.
Find the molar solubility of a given compound in the specific solution with the help of expression of solubility product.
Solubility product:
Solubility product basically defines the equilibrium between solids and its respective ions in the solution. The value of the solubility product represents the degree to which a chemical compound dissociates in water.
Example –
The dissociation reaction of ionic solid 
is given as shown below.
The expression corresponding to the solubility product is given as shown below.
![K. =[A+][B]](http://img.homeworklib.com/questions/0f331550-0d22-11ea-a629-1bf3daa0fcd2.png?x-oss-process=image/resize,w_560)
The balanced chemical equation for the dissociation reaction of 
is as follows:
The expression corresponding to the solubility product is given as shown below.
![Ko =[Ag+ ][Br]](http://img.homeworklib.com/questions/1054e090-0d22-11ea-8ce7-790acf77746e.png?x-oss-process=image/resize,w_560)
…… (1)
The molar solubility of is calculated as shown below.
Consider the 
is the molar solubility of the given compound.
Substitute 
as the concentration of bromide ion and 
as the 
of in the equation 
.
The molar solubility of 
is 
.
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Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
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Determine the molar solubility of AgBr in a solution containing 0.250 M MgBr2 (strong electrolyte). Ksp (AgBr) = 7.7 x 10-13 O A.8.8 x 107 M B.5.1 x 10-12 M c. 1.5 x 10-12 M O 0.5.8 x 10-5 M E.0.150 M
The
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