1.Calculate the pH needed to start precipitation of tin (II) sulfide from a solu

Question

1.Calculate the pH needed to start precipitation of tin (II) sulfide from a solution containing 0.0049 M of the tin (II) ion. The sulfide ion is produced in a solution saturated with H2S via the following equilibrium: [H+]2 [S2] = 1.1 x 1022. (Ksp = 1.3 x 1023 for tin (II) sulfide)
1.Calculate the pH needed to start precipitation of tin (II) sulfide from a solution containing 0.0049 M of the tin (II) ion. The sulfide ion is produced in a solution saturated with H2S via the following equilibrium: [H+]2 [S2] = 1.1 x 1022. (Ksp = 1.3 x 1023 for tin (II) sulfide)
1.Calculate the pH needed to start precipitation of tin (II) sulfide from a solution containing 0.0049 M of the tin (II) ion. The sulfide ion is produced in a solution saturated with H2S via the following equilibrium: [H+]2 [S2] = 1.1 x 1022. (Ksp = 1.3 x 1023 for tin (II) sulfide)

Explanation / Answer

SnS --------------------> Sn+2 + S^-2

Ksp = [Sn+2] [S^-2]

1.3 x 10^-23 = 0.049 x (S-2)

[S-2] = 2.65 x 10^-22 M

[H+]2 [S2] = 1.1 x 1022

[H+]^2 x 2.65 x 10^-22 = 1.1 x 1022

[H+] = 0.643 M

pH = -log [H+]

pH = -log(0.643)

pH = 0.19

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