Calculate the solubility of AgCl(s) in pure water. Given solubility product of AgCl = 2.8 X 10-10 .
Answers
Answer:
The chloride ion concentration in 0.01M HCl will be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
K
sp
=[Ag
+
][Cl
−
]
Substitute values in the above expression.
2.8×10
−10
=[Ag
+
]×0.01
Hence, [Ag
+
]=
0.01
2.8×10
−10
=2.8×10
−8
mol/L.
YOUR ANSWER IS : 2.8 X 10-⁸ mol/ L
The chloride ion concentration in 0.01M HCl will
be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
K sp =[Ag +][Cl − ]
Substitute values in the above expression.
2.8×10 −10 =[Ag + ]×0.01
Hence,
[Ag + ]= 0.012.8×10 −10
.. =2.8×10 −8 mol/L.
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