Hello, I have the next problem about solubility and dissociation of ions:
Calculate the solubility of FeS in pure water. Ksp = 8 × 10−19. (Hint: The second stage of hydrolysis, producing
H2S, cannot be neglected.)
This excersice was taken from Schaum's college chemistry book and the answer has to be 4,0×10^-7.
In the book is also mentioned the constants of dissociation for H2S (the acid)
Ka1 = 1,0×10^-7
Ka2 = 1,2 ×10^-13
I really don't know how to get the answer, Please help
2 answers
this mite help you: go to mathway
Mmm well I'm doing using the mathway but I am a bit confused.
I have six equations by the way:
Ksp= [Fe][S] (eq.1 from solub.)
Ka1= [HS]×[H3O]/[H2S] (eq.2 from first dissociation of the acid)
ka2= [S]×[H3O]/[HS] (eq. 3 from second dissociation)
Kw= [H3O][OH] (eq. 4 from water ionization, assuming that it will affect the concentration of (OH))
[Fe]=[S]+[HS]+[H2S] (eq.5 from a mass balance)
2[Fe]+[H3O]=2[S]+[HS]+[OH](eq. 6 from charge balance)
Now I am strugglin about how to resolved the system of equations, I was thinking about simplifying some variables in 5 or 6 but I don't know what it would be...
Could you help me in this?
I have six equations by the way:
Ksp= [Fe][S] (eq.1 from solub.)
Ka1= [HS]×[H3O]/[H2S] (eq.2 from first dissociation of the acid)
ka2= [S]×[H3O]/[HS] (eq. 3 from second dissociation)
Kw= [H3O][OH] (eq. 4 from water ionization, assuming that it will affect the concentration of (OH))
[Fe]=[S]+[HS]+[H2S] (eq.5 from a mass balance)
2[Fe]+[H3O]=2[S]+[HS]+[OH](eq. 6 from charge balance)
Now I am strugglin about how to resolved the system of equations, I was thinking about simplifying some variables in 5 or 6 but I don't know what it would be...
Could you help me in this?
2 answers