Using Appendix D in the textbook, calculate the molar solubility of AgBr in 3.1×10−2 M AgNO3 solution. Silver bromide Ag...

Question

Question

Using Appendix D in the textbook, calculate the molar solubility of AgBr in 3.1×10⁻² M AgNO3 solution.

Silver bromide AgBr 5.0 * 10-13

science chemistry

Answer 1

Answer #1

AgNO3 here is Strong electrolyte

It will dissociate completely to give [Ag+] = 0.031 M

At equilibrium:

AgBr <----> Ag+ + Br-

3.1*10^-2 +s s

Ksp = [Ag+][Br-]

5*10^-13=(3.1*10^-2 + s)*(s)

Since Ksp is small, s can be ignored as compared to 3.1*10^-2

Above expression thus becomes:

5*10^-13=(3.1*10^-2)*(s)

5*10^-13= 3.1*10^-2 * 1(s)^1

s = 1.613*10^-11 M

Answer: 1.6*10^-11 M

Answer 2

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