1. 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 =
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 = = log [(NO3-) / (NO2-)] – 2(pH + pE)
-28.2 = log [(NO3-) / (NO2-)] – 26.0
-2.2 = log [(NO3-) / (NO2-)] So, [(NO3-) / (NO2-)] =

2. 11.
1/8 SO /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 – pH
1/8 SO /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

3. 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-] = pE + 5/4 pH + log [H2S] = -5.1 – – 2 = -5.1
at pE = -3, pH = 4 and log [S]T = -2 = log [H2S] = -2.

4. 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 SO /8 H+ + e = 1/8 HS- + 1/2 H2O log K = 4.3
1/8 HS- + 1/8 H+ = 1/8 H2S log K =
1/8 SO /8 H+ + e = 1/8 H2S + 1/2 H2O log K =
pE = – 1.25 pH
Plot red line from pH = 0, pE = to pH = 7, pE =
Plot black line as done except from pH = 7, pE =