Determine the molar solubility of AgBr in a solution at 25°C containing 0.0005 M

Question

Determine the molar solubility of AgBr in a solution at 25°C containing 0.0005 M AgNO3 AND say whether AgBr is more or less soluble in this solution versus in pure water. The Ksp of AgBr at 25°C is 7.7x10-13.

a. 7.5x10-10 M, AgBr is more soluble in this solution than in pure water

b. 7.5x10-10 M, AgBr is less soluble in this solution than in pure water

c. 1.5x10-9 M, AgBr is more soluble in this solution than in pure water

d. 1.5x10-9 M, AgBr is less soluble in this solution than in pure water

e. 3.9x10-16 M, AgBr is more soluble in this solution than in pure water

f. 3.9x10-16 M, AgBr is less soluble in this solution than in pure water

Explanation / Answer

given AgBr

AgBr ---> Ag+ + Br-

let the molar solubility of AgBr be s

now

in pure water

[Ag+] = [Br-] = s

now

Ksp = [Ag+] [Br-]

Ksp = [s] [s]

Ksp =s2

s2 = 7.7 x 10-13

s = 8.775 x 10-7

so

the molar solubility of AgBr in pure water is 8.775 x 10-7 M

2)

now in 0.0005 M AgN03

AgN03 ---> Ag+ + N03-

we know that

AgN03 is completely soluble

so

[Ag+] = 0.0005

now

AgBr ---> Ag+ + Br-

let the molar solubility of AgBr be s

then

[Br-] = s

but due to common ion effect

[Ag+] = 0.0005

now

Ksp = [Ag+] [Br-]

7.7 x 10-13 = [0.0005] [s]

s = 1.54 x 10-9 M

so

the molar solubility of AgBr in AgN03 solution is 1.54 x 10-9 M

and

we got that

the molar solubility of AgBr in pure water is 8.775 x 10-7 M

so

we can see that

AgBr is less soluble in the solution than in pure water

so

the answer is

d) 1.5x10-9 M, AgBr is less soluble in this solution than in pure water

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