Question

calculate solubility of agcl in 0.1 molar NaCl at 25 degree Celsius if the solubility product at same temperature is 2 into 10 raise to minus 10

Answer 1

agcl=ag+ + cl-
initial moles of agcl is 's'
after dissociation ag+ =s. and. cl-=s
and,
given that,
Nacl =0.2M
Nacl=Na+ +cl-
so, Na+ =0.2
and cl- =0.2
cl- =s+0.2
ksp=[0.2+s][s]
=assume that s<<0.2 so it is neglected
2*10^-10=0.2.s
s=2*10^-10/0.2
I think, you can do calculations
s=10^-9

Answer 2

Answer:

this is the best i can di