Treatment of alkaline wastewater
Thread Content
There is a mixed waste liquid containing approximately 2.5% NaOH, about 2.0% Al2O3, and around 0.6% Na2C2O4; it also contains small amounts of sodium carbonate and other trace impurities. It is necessary to separate Na2C2O4 from this mixture. What are some good methods for treating such wastewater?
What is Na2C2O4? Could it be that I wrote it wrong?
That’s sodium oxalate, the simplest binary organic salt.
There is no good method to separate sodium oxalate from carbon steel in an alkaline solution, as both have high water solubility, and separation is difficult even after concentration. Additionally, sodium oxalate also decomposes easily when heated. If there is a cheap source of oxalic acid, neutralizing the alkaline solution with oxalic acid can yield a solution containing mainly sodium oxalate, which may make separation easier at this point. The cost of oxalic acid is also not low; you need to calculate the costs and benefits.
There is an alkaline solution containing about 2.5% NaOH and about 2.0% Al2O3; Al2O3 + 2NaOH == 2NaAlO2 (sodium aluminate) + H2O. When CO2 gas is introduced into the NaAlO2 solution, if there is a small amount of CO2 initially, the following reaction occurs: 2NaAlO2 + CO2 + 3H2O = 2Al(OH)3↓ + Na2CO3. As more CO2 is introduced, the excess CO2 reacts with the Na2CO3 produced in the previous reaction to yield: 2CO2 + H2O + Na2CO3 == 2NaHCO3. Oxalic acid is then added; if there is an excess of oxalic acid, the reaction is: 2H2C2O4 + Na2CO3 = CO2 + H2O + 2NaHC2O4. If there is a shortage of oxalic acid, the reaction is: H2C2O4 + 2Na2CO3 = 2NaHCO3 + Na2C2O4. Finally, evaporation is carried out. It is theoretically feasible; it depends on whether the costs and benefits make it worthwhile. Provided by Kaifeng Hongyuan Water Treatment.
Hello, could you be more specific? How to handle NaHCO3? How is the alkali in water recovered?
CO2 (in small amounts) + 2NaOH = Na2CO3 + H2O; CO2 (in excess) + NaOH = NaHCO3. There is no more base left in the water. H2C2O4 + 2NaHCO3 = Na2C2O4 + 2H2O + 2CO2. In the end, only Na2C2O4 and water remain, and evaporation can be used to separate them. This is the theory; it’s best for you to have a laboratory verify it first
Hello, our goal is not to remove NaOH; in alkaline solutions, it is necessary to recover NaOH, and it is sufficient to only remove the impurity Na2C2O4.