7. ½ NO2- + ½ H2O = ½ NO3- + H+ + e log K = -14.1 ¼ O2 + H+ = ½ H2O log K = 20.8 Which sum to give, ½ NO2- + ¼ O2 = ½ NO3- log K = 6.7 With an overall reaction and log K of, NO2- + ½ O2 = NO3- log K = 13.4 From the 1st ½-reaction multiplied by 2, log K = -28.2 = log [(NO3-) / (NO2-)] – 2(pH + pE) -28.2 = log [(NO3-) / (NO2-)] – 26.0 -2.2 = log [(NO3-) / (NO2-)] So, [(NO3-) / (NO2-)] = 10-2.2

11. 1/8 SO42- + 9/8 H+ + e = 1/8 HS- + 1/2 H2O log K = 4.3 log K = 1/8 log (HS-) – 1/8 log (SO42-) + 9/8 pH + pE pE = 4.3 – 1.125 pH 1/8 SO42- + 5/4 H+ + e = 1/8 H2S + 1/2 H2O log K = 5.1 log K = 1/8 log (H2S) – 1/8 log (SO42-) + 5/4 pH + pE pE = 5.1 – 1.25 pH Add the reverse of the first reaction to the second reaction to get 1/8 HS- + 1/8 H+ = 1/8 H2S log K = 0.8 HS- + H+ = H2S log K = 6.4

oxidized SO42- and black separates HS- from SO42-. Upper and lower bold blue lines are limits of stability of water. See handout. vertical segment separates H2S from HS-. Red separates reduced H2S from oxidized SO42- and black separates HS- from SO42-. If specify total S concentration, can estimate concentration of minor species in field where major species dominates. For example, log [SO42-] = -5.1 + pE + 5/4 pH + log [H2S] = -5.1 – 3 + 5 – 2 = -5.1 at pE = -3, pH = 4 and log [S]T = -2 = log [H2S] = -2.

Some chose to use a literature value for K = (HS-)(H+) / (H2S) = 10-7. This leads to an inconsistency unless one or the other of the equations given in Table 6.2 are adjusted. For example, 1/8 SO42- + 9/8 H+ + e = 1/8 HS- + 1/2 H2O log K = 4.3 1/8 HS- + 1/8 H+ = 1/8 H2S log K = 0.875 1/8 SO42- + 10/8 H+ + e = 1/8 H2S + 1/2 H2O log K = 5.175 pE = 5.175 – 1.25 pH Plot red line from pH = 0, pE = 5.175 to pH = 7, pE = -3.575. Plot black line as done except from pH = 7, pE = -3.575.