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+(aq) + Br-(aq) 0.03 (if x is very small)[Ag+] = x = 1.67 x 10-11 M
Answered by
| 4th Jun, 2014,
03:23: PM
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