Alkaline Activated Persulfate
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You just completed a bench test to evaluate alkaline acitvated persulfate for in situ application at your site. The results indicate that Soil Oxidant Demand (SOD) is low, chemicals of concern (COCs) are destroyed, and no hexavalent chromium was generated. You want to use this technology. How much sodium persulfate (SP) and sodium hydroxide (NaOH) do you need?
PRIMA suggests the following approach:
- Calculate the mass of SP based on measured SOD and COC concentrations
- Compare the concentration this SP will generate in GW to the SP concentration used in the COC Removal test. Increase SP if needed to improve COC removal.
- Calculate the NaOH requirement based on the mass of SP and initial buffering ability of soil and GW.
- Compare this amount to the amount of NaOH buffered by soil in the long term soil buffering curve test. If necessary, increase NaOH to ensure soil alone does not neutralize the added NaOH.
The total mass of SP required (SPtotal) is based on the measured SOD and on the concentration of COCs in soil and groundwater at the site; that is
SPtotal = SPSOD + SPCOCs (1)
where SPSOD is the mass due to SOD and SPCOCs is the mass due to COCs. For a soil with the properties given in Table 1,
SPSOD = 2.2 g SP/kg x 1,800 kg = 3,960 g
The theoretical amount of SP needed to convert TCE to carbon dioxide is 5.4 g SP/gTCE, so SP due to COCs is
SPCOCs = 5.4 g SP/g TCE x [(0.016 g TCE/kg x 1,800 kg) + (0.0012 g TCE/L x 300L)] = 157 g SP
SPtotal = 3,960 g + 157 g = 4,120 g SP/m3 soil
The concentration of SP after injection would be 4,120 g/300L = 13.7 g SP/L. This value should be compared to the COC Removal data generated during the bench test and adjusted up, if needed, to ensure that COCs can be destroyed.
Table 1. Example Soil Properties
|Volume soil||1 m3 (1000 L)|
|Mass soil||1,800 kg|
|Volume water||300 L|
|SOD||2.2 g SP/kg soil|
|Initial Buffering||0.038 mol OH-/kg soil
0.012 mol OH-/L GW
|COCs (TCE)||1,200 μg/L in GW
16,000 μg/kg in soil
Once the mass of SP is known, the amount of NaOH can be easily calculated. Two moles of NaOH per mole SP are needed to neutralize the acid generated by decomposition of SP. In addition, soil can buffer against changes in pH, so “extra” NaOH is required. The total NaOH requirement is thus
NaOHtotal = NaOHSP + NaOHbuf (2)
where NaOHSP is the moles due to SP and NaOHbuf is moles due to the initial buffering of soil and GW. For the soil in Table 1,
NaOHSP = 4,120 g SP x (1 mole SP/238 g SP) x 2 mole NaOH/mole SP = 35 moles NaOH
The amount of NaOH to overcome the initial buffering of soil and GW is
NaOHbuf = (0.038 mole/kg soil x 1,800 kg) + (0.012 mole/L GW x 300 L) = 72 moles NaOH
NaOHtotal = 35 moles + 72 moles = 107 moles NaOH/m3soil = 59 millimoles NaOH/kg soil
This result should be compared to the long-term soil buffering curves, examples of which are shown in Figure 1. Soil A buffers poorly – pH did not attenuate upon exposure to 100 mmol OH-/kg soil. Thus, 59 mmoles NaOH/kg soil should enable the pH to remain above 10.5 (the level for SP activation) until the NaOH is neutralized via SP decomposition. In contrast, Soil B buffers well. The pH attenuated from 13 to < 10.5 within 7 days upon exposure of to the same amount of NaOH, so more than 59 mmole/kg NaOH may be needed to maintain adequate pH.
Figure 1. Example Soil Buffering Curves