Calculate [Ag+], when 0.03 mole of KBr is added to one liter saturated solution of AgBr. Given Ksp of AgBr =5.0x10-13

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

                     AgBr(s)   Ag+(aq) + Br-(aq)
                        
                        Ksp = [Ag+][Br-]=5.0x10-13       
 
If the equilibrium concentration of the silver ion is x, then the equilibrium concentration of the bromide ion is (x + 0.030).
 
x + 0.030  0.03 (if x is very small)
 
or,   Ksp = [Ag+][Br-] = 5.0 x 10-13 = (0.030)(x)
 

[Ag+] = x = 1.67 x 10-11 M

Answered by  | 4th Jun, 2014, 03:23: PM