Solve Ksp Equation for Sn^2+ When MnS Begins to Precipitate

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In summary, for stannous sulfide, SnS, the Ksp is 1.0X10^-25 and for manganese sulfide, MnS, the Ksp is 3.0 x 10^-15. The question asks for the molar concentration of Sn^(2+) ion when manganese sulfide first starts to precipitate in a solution containing .01 M Sn^2+ and .01 M Mn^2+ with increasing sulfide ion concentration. After a logical progression, the molar concentration of Sn^(2+) is calculated to be 3.33E-9M, but the correct answer is 3.33E-13M.
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Newtime
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Homework Statement



For stannous sulfide, SnS, the Ksp = 1.0X10^-25
For manganese sulfide, MnS, the Ksp = 3.0 x 10^-15

Assume a solution contains .01 M Sn^2+ and .01 M Mn^2+. If the sulfide ion (S^2-) concentration is increased gradually without any dilution of the first solution, what will be the molar concentration of Sn^(2+) ion when manganese sulfide first starts to precipitate?

Homework Equations



Ksp = ?

The Attempt at a Solution



I've tried this question in what I, and a few of my friends who are stuck on it with me, consider a logical progression. I ended up with [Sn] = 1E-25, [Mn] = 3E-15, =1E-27 and =3E-17, respectively. [Sn][3E-17]=1E-15 and I end up with [Sn]=3.33E-9M. The answer is 3.33E-13 and I don't understand how it was attained. Any help is appreciated, thank you.
 
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  • #2
0.01x10-17 is not 10-15.
 
  • #3
Borek said:
0.01x10-17 is not 10-15.

Which part of my post are you referring to?
 
  • #4
Anyone. You did the same mistake several times.
 

1. What is the Ksp equation for Sn^2+ when MnS begins to precipitate?

The Ksp equation for Sn^2+ when MnS begins to precipitate is Ksp = [Sn^2+][S^2-]. This represents the solubility product constant for the salt SnS, which is formed when Sn^2+ ions combine with S^2- ions to form a precipitate.

2. How do you calculate the Ksp for Sn^2+ when MnS begins to precipitate?

To calculate the Ksp for Sn^2+ when MnS begins to precipitate, you need to know the molar solubility of the salt MnS and the concentration of Sn^2+ ions in the solution. The Ksp can be determined by multiplying the molar solubility of MnS by the concentration of Sn^2+ ions, as represented in the Ksp equation.

3. What is the significance of MnS precipitating in the presence of Sn^2+ ions?

MnS precipitating in the presence of Sn^2+ ions indicates that the solubility product of SnS has been exceeded, causing the formation of a solid precipitate. This is important because it allows for the determination of the Ksp for Sn^2+ and can also be used to understand the solubility behavior of SnS in different solutions.

4. How does the concentration of Sn^2+ ions affect the formation of MnS precipitate?

The concentration of Sn^2+ ions directly affects the formation of MnS precipitate. As the concentration of Sn^2+ ions increases, the likelihood of the formation of SnS and precipitation of MnS also increases. This is because more Sn^2+ ions are available to react with S^2- ions and form the insoluble salt MnS.

5. Can the Ksp value for Sn^2+ be used to predict the solubility of other salts containing Sn^2+ ions?

Yes, the Ksp value for Sn^2+ can be used to predict the solubility of other salts containing Sn^2+ ions. This is because the Ksp value represents the equilibrium constant for the formation of SnS, which is a common component in many salts containing Sn^2+ ions. However, other factors such as pH and temperature may also affect the solubility of these salts.

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