Question

In the Group 3 precipitations, the pH is raised by adding Ammonia. Suppose it is raised to pH = 9. What is the [S2- ] concentration under these conditions? Assume the H2S concentration is again 0.1M.

Consult an appropriate source and determine the Concentration of each of the following species, when in Concentrated form: HCl, H2SO4, and NH3.

Answer 1

In the Group 3 precipitations, the pH is raised by adding Ammonia. Suppose it is raised to pH = 9. What is the [S2- ] concentration under these conditions? Assume the H2S concentration is again 0.1M.Consult an appropriate source and determine the Concentration of each of the following species, when in Concentrated form: HCl, H2SO4, and NH3

solution:

These Cations precipitate in an Ammoniacal Solution. Because Ammonia solutions are Basic (OH- ):

H2S(aq) <------> H+(aq) + HS-(aq)

K1 = [H+][HS-] / [H2S] = 1.0x10^-7

HS-(aq) <------> H+(aq) + S2-(aq)

K2 = [H+][S2-] / [HS-] = 1.3x10^-13

Since K2 is much smaller compared to K1, the H+ concentration is determined by the first step.

pH = 9
[H+] = 10^-pH = 10^-9 = 1x10^-9 M

H2S(aq) <------> H+(aq) + HS-(aq)
0.10 - x M ......... 1x10^-9 M ... x M

K1 = [H+][HS-] / [H2S]

after calculateing the value of x
x = 0.0999 M = [HS-]

HS-(aq) <------> H+(aq) + S2-(aq)
0.0999 - y M ... 1x10^-9M ... y M

K2 = [H+][S2-] / [HS-]
(1x10^-9)(y) / (0.0999)
y = 0.0998=[S2-]

[H2S] = 0.10 - x M = 0.10 - 0.0999 = 1 x 10^4 M
[H3O1+] = 1x10^-9 M
[HS1-] = 0.0999M
[S2-] = 0.0998M