3) Given that the standard free energies of formation of Ag (aq),
Cl(aq), and AgCl(s) are 77.1 kJ/mol, -131.2 kJ/mol, and
-109.8 kJ/mol, respectively, calculate the solubility product,
Ksp, for AgCl.
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0
Answer:
Correct option is A)
ΔG=ΔG(AgCl(s))−[ΔG(Ag
+
)+ΔG(Cl
−
)]=−109kJ/mol−[77kJ/mol−129kJ/mol]=−57kJ/mol
ΔG=−nFE
E=
nF
−ΔG
=
1×96500
−57000J/mol
=0.591V
E
0
=0.222V
E=E
0
−
n
0.0592
log
[Ag
+
][Cl
−
]
1
0.59V=0.222V−
1
0.0592
log
K
sp
1
log
K
sp
1
=6.2276
K
sp
1
=1.688×10
6
K
sp
=5.9×10
−7
The molar mass of AgCl is 143.5 g/mol.
K
sp
=
143.5
5.9×10
−7
=4×10
−9
Hence, K
sp
×10
10
≃40
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