Sodium thiosulfate (STS) is a chemical compound used to neutralize chlorine and chloramines in water. It acts as an effective dechlorinating agent across various applications, including aquaculture, swimming pool maintenance, and preparing samples for drinking water analysis. Understanding precisely how much sodium thiosulfate is needed for complete neutralization is fundamental. The required amount is based on a specific chemical relationship that translates into practical dosing calculations.
The Chemical Reaction: Neutralizing Chlorine
Neutralization involves a chemical change where toxic chlorine is converted into benign substances. When chlorine, often present as hypochlorous acid or hypochlorite ions, encounters sodium thiosulfate, a reduction-oxidation reaction occurs. Sodium thiosulfate acts as a reducing agent, donating electrons to the chlorine compounds, which transforms the free chlorine into stable, harmless chloride ions.
During this reaction, the sulfur component of the thiosulfate is oxidized into other sulfur-containing compounds. Depending on the \(\text{pH}\) and concentration, the final sulfur product is typically either tetrathionate or sulfate. This conversion effectively removes the oxidizing and disinfecting properties of the chlorine, making the water safe for aquatic life or other sensitive applications.
Determining the Stoichiometric Ratio
The required amount of sodium thiosulfate is determined by the stoichiometric ratio, the precise mole-to-mole relationship established by the balanced chemical equation. For the main reaction with free chlorine, the theoretical minimum ratio is approximately one mole of sodium thiosulfate to neutralize four moles of chlorine. Translating this molar ratio into a practical mass ratio requires accounting for the molecular weights of both chlorine and sodium thiosulfate.
Sodium thiosulfate is sold in two common forms: anhydrous (\(\text{Na}_2\text{S}_2\text{O}_3\)) and pentahydrate (\(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}\)). The pentahydrate form is approximately 1.57 times heavier than the anhydrous form for the same amount of active thiosulfate. The theoretical minimum for the anhydrous form is roughly 1.4 milligrams of STS per 1 milligram of chlorine, meaning the pentahydrate form requires approximately 2.2 milligrams.
In practice, a higher ratio is always used to ensure complete and rapid neutralization. For reliable dechlorination, a practical dosing ratio of 5 to 7 parts of sodium thiosulfate pentahydrate to 1 part of chlorine is commonly recommended. This safety margin accounts for variations in water chemistry, \(\text{pH}\), and the presence of chloramines. A safe, practical guideline is to use 7 milligrams of sodium thiosulfate pentahydrate for every 1 milligram of chlorine present.
Practical Dosing Calculations for Common Scenarios
Translating the stoichiometric ratio into a usable amount involves converting the chlorine concentration, typically measured in parts per million (\(\text{ppm}\)), into a total mass of chlorine within the volume of water being treated. The required mass of sodium thiosulfate is found by multiplying this total chlorine mass by the chosen practical dosing ratio.
For example, when treating municipal tap water for an aquarium, the concentration is usually around \(1 \text{ ppm}\). If 100 liters of water contains \(1 \text{ ppm}\) of chlorine, this equals \(100 \text{ mg}\) of chlorine to neutralize. Using the practical 7:1 ratio for the pentahydrate form, you would need \(700 \text{ mg}\) (or \(0.7\) grams) of sodium thiosulfate pentahydrate. This ensures the low-level chlorine is quickly and fully removed.
For applications involving higher concentrations, such as a shock-treated pool, the calculation remains the same but the scale increases significantly. If a 10,000-gallon swimming pool needs its free chlorine level lowered by \(5 \text{ ppm}\), the total mass of chlorine to be neutralized is substantial. Since 10,000 gallons is approximately 37,850 liters, a \(5 \text{ mg/L}\) reduction means \(189,250 \text{ mg}\) of chlorine must be removed.
Applying the 7:1 ratio means you would need approximately \(1.325\) kilograms of sodium thiosulfate pentahydrate. Since the anhydrous form is more concentrated, using it requires a mass approximately 1.57 times less than the pentahydrate. Accurately knowing the form of sodium thiosulfate being used is paramount for correct dosing.
Safety and Handling of Sodium Thiosulfate
When handling sodium thiosulfate, particularly the dry, crystalline forms, users should avoid direct contact with the powder and solutions. Appropriate personal protective equipment, such as gloves and chemical safety goggles, must be worn. Adequate ventilation should be maintained, especially when mixing large batches, to prevent the inhalation of dust or mist.
Sodium thiosulfate should be stored in a cool, dry, and well-ventilated location, kept away from incompatible substances like strong acids and oxidizing agents. The pentahydrate form is more stable and easier to handle than the anhydrous form, which can be hygroscopic and absorb moisture. When preparing solutions, dissolving the anhydrous form can generate heat, while the pentahydrate form will cool the solution.
Caution is advised against excessive overdosing in aquatic environments. Adding a grossly oversized amount can lead to the consumption of dissolved oxygen in the water as the excess thiosulfate oxidizes over time. This oxygen depletion can be harmful to fish and other aquatic organisms, reinforcing the need for accurate and measured dosing.