Pond Eutrophication: Phosphate Limitation and an Alum Dose Decision
A citizen-science group monitors a 2.0 × 10^5 L farm pond. Upstream water: 0.80 mg/L NO3−-N, 0.05 mg/L PO4^3−-P, DO 8.1 mg/L. Downstream, after a tile drain: 8.5 mg/L NO3−-N, 0.45 mg/L PO4^3−-P, DO 3.2 mg/L. Chlorophyll a is 38 µg/L (upstream 4 µg/L). The group proposes dosing the pond with alum, Al2(SO4)3·18H2O, to precipitate phosphate as AlPO4. A member calculates that 90 g of phosphate-P must be removed and wants to add 1.0 kg of alum to be safe. The pond has alkalinity 45 mg/L as CaCO3 and pH 7.8. Residents swim in the pond, and a fish kill was reported last summer. The group has a $200 budget, a conductivity meter, and no laboratory. They can collect grab samples and send them to a county lab only once.
1. Identify the likely limiting nutrient for the algal bloom and justify with the provided concentrations. Explain why the dissolved oxygen drop matters.
Read the worked answer
Compare N and P on a molar basis. Downstream NO3−-N = 8.5 mg/L ÷ 14.0 g/mol = 0.607 mmol/L; PO4-P = 0.45 mg/L ÷ 31.0 g/mol = 0.0145 mmol/L; molar N:P ≈ 42:1. The Redfield ratio for algal biomass is about 16:1, so there is far more N than P relative to demand; P is the nutrient in shortest supply and is therefore the likely limiting nutrient. Upstream N:P is also high (≈35:1), and the bloom appears after the P increase, consistent with P control.
The DO drop from 8.1 to 3.2 mg/L indicates that respiration/decomposition of algal organic matter consumes oxygen faster than reaeration; low DO stresses fish and can cause the reported fish kill. Therefore reducing P is a logical target, but the oxygen problem also requires addressing the organic load and circulation.
2. Calculate the minimum mass of Al2(SO4)3·18H2O needed to precipitate the phosphate in the pond, assuming 1:1 Al:PO4 and no competing reactions. Compare with the proposed 1.0 kg and identify two reasons the real dose may differ.
Read the worked answer
Phosphate-P to remove: 0.45 mg/L × 2.0×10^5 L = 9.0×10^4 mg = 90 g P. Moles P = 90 g ÷ 30.97 g/mol = 2.91 mol. Each AlPO4 requires one Al, and each Al2(SO4)3·18H2O supplies 2 Al, so alum needed = 2.91/2 = 1.45 mol. Molar mass of the hydrate ≈ 2(26.98) + 3(32.07) + 12(16.00) + 18(18.02) = 666.4 g/mol; mass = 1.45 × 666.4 = 966 g ≈ 0.97 kg.
The proposed 1.0 kg is only about 3–4% above the stoichiometric minimum, not a safety margin. Real dose may differ because (1) Al3+ also reacts with OH−/alkalinity to form Al(OH)3, consuming alum and lowering pH; (2) not all phosphate is dissolved or reactive, and mixing, competing ligands, and sediment release affect contact; (3) overdosing can leave soluble Al and harm fish. A jar test and alkalinity/pH monitoring are needed.
3. Recommend whether the group should proceed with the alum treatment, and outline a monitoring/decision plan. Include one alternative or complementary action.
Read the worked answer
Do not add 1.0 kg directly to the pond on this evidence. The dose is near-stoichiometric with little margin, the pond has only 45 mg/L alkalinity as CaCO3, and alum can lower pH and mobilize Al3+, which is toxic to fish; the prior fish kill and swimming use raise the stakes.
First run a bench jar test on pond water with a range of alum doses, measure pH and turbidity/conductivity, and check whether phosphate actually falls. If dosing proceeds, add in a well-mixed area, not near the entire pond, and monitor pH, DO, clarity, and phosphate before, during, and after; have a stop threshold for pH or Al. The group can send one grab sample to the county lab for confirmation. A complementary action is source control: a vegetated buffer strip or controlled drainage to reduce P input, plus aeration or a barley-straw trial only with monitoring. If alkalinity is insufficient, do not use alum without a buffer.