Doobwa
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Hello! I've just completed this homework problem but got ALL answers incorrect. I really thought I knew what I was doing so I'm kind of surprised. Could someone help point out what I had done wrong?
Ag2CO3, is considerably more soluble in hot than in cold water. At 20 °C, 0.0032 g of this compound dissolves in 0.100 L of water, but 0.050 g dissolves at 100.0 °C. Calculate Ksp and G° in kJ/mol at each temperature.
Ksp=[Ag[itex]^{+}[/itex]][itex]^{2}[/itex][CO3[itex]^{-}[/itex]]
deltaG=-RTlnKsp.
Solubility=(mass)/(volume)(molar mass)
I think I went wrong with figuring out Ksp. I believe if I knew what I had done wrong here, I could easily solve the rest, so I'll just show my work for this portion.
At 20C (293.15K) 0.0032g of Ag2CO3 dissolves in 0.100L of water. Molar mass of Ag2CO3 found to be 275.75g/mol.
Solubility of Ag2CO3= (0.0032g)/(0.100L)(275.75g/mol)=1.160471442e-4M
Ag2CO3[itex]\searrow[/itex]2Ag[itex]^{+}[/itex]+CO3[itex]^{-}[/itex]
[Ag[itex]^{+}[/itex]]=2.320942884e-4M
[CO3[itex]^{-}[/itex]]=1.160471442e-4M
Ksp=[Ag[itex]^{+}[/itex]][itex]^{2}[/itex][CO3[itex]^{-}[/itex]]
=(2.320942884e-4)[itex]^{2}[/itex](1.160471442e-4)
=6.25e-12
Actual answer: 2.77e-11
I really appreciate any guidance you can give me!
Homework Statement
Ag2CO3, is considerably more soluble in hot than in cold water. At 20 °C, 0.0032 g of this compound dissolves in 0.100 L of water, but 0.050 g dissolves at 100.0 °C. Calculate Ksp and G° in kJ/mol at each temperature.
Homework Equations
Ksp=[Ag[itex]^{+}[/itex]][itex]^{2}[/itex][CO3[itex]^{-}[/itex]]
deltaG=-RTlnKsp.
Solubility=(mass)/(volume)(molar mass)
The Attempt at a Solution
I think I went wrong with figuring out Ksp. I believe if I knew what I had done wrong here, I could easily solve the rest, so I'll just show my work for this portion.
At 20C (293.15K) 0.0032g of Ag2CO3 dissolves in 0.100L of water. Molar mass of Ag2CO3 found to be 275.75g/mol.
Solubility of Ag2CO3= (0.0032g)/(0.100L)(275.75g/mol)=1.160471442e-4M
Ag2CO3[itex]\searrow[/itex]2Ag[itex]^{+}[/itex]+CO3[itex]^{-}[/itex]
[Ag[itex]^{+}[/itex]]=2.320942884e-4M
[CO3[itex]^{-}[/itex]]=1.160471442e-4M
Ksp=[Ag[itex]^{+}[/itex]][itex]^{2}[/itex][CO3[itex]^{-}[/itex]]
=(2.320942884e-4)[itex]^{2}[/itex](1.160471442e-4)
=6.25e-12
Actual answer: 2.77e-11
I really appreciate any guidance you can give me!