Calculate the molar solubility in pure water for the following compound at 25Â°C, Ag2SO3 Ksp = 1.5x10-14 What...

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Calculate the molar solubility in pure water for the followingcompound at 25Â°C, Ag₂SO₃ K_sp =1.5x10^-14
What is the molar solubility in 0.0010 M sodiumsulfite?Â Â

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4.5 Ratings (889 Votes)

1)
At equilibrium:
Ag2SO3Â Â Â <---->Â Â Â 2 Ag+Â Â Â Â +Â Â Â Â Â Â Â Â Â SO32-Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â sÂ Â Â Â Â Â Â

Ksp = [Ag+]^2[SO32-]
1.5*10^-14=(2s)^2*(s)
1.5*10^-14= 4(s)^3
s = 1.554*10^-5 M
Answer: 1.6*10^-5 M

2)
Na2SO3 here is Strong electrolyte
It will dissociate completely to give [SO32-] = 0.001 M
At equilibrium:
Ag2SO3Â Â Â <---->Â Â Â 2 Ag+Â Â Â Â +Â Â Â Â Â Â Â Â Â SO32-Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1*10^-3 + s
Ksp = [Ag+]^2[SO32-]
1.5*10^-14=(2s)^2*(1*10^-3+ s)
Since Ksp is small, s can be ignored as compared to 1*10^-3
Above expression thus becomes:
1.5*10^-14=(2s)^2*(1*10^-3)
1.5*10^-14= 4(s)^2 * 10^-3
s = 1.936*10^-6 M
Answer: 1.9*10^-6 M

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