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# Calculating Necessary Tonnes of Lime to Raise a Lake's pH

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One of the impacts of fossil fuels is an increase in acid deposition (or acid rain as many people refer to it). The following exercise asks you to calculate how much lime (an alkaline rock, not the tasty green fruit) it would take to increase the pH of a small lake.

Consider a small lake in the Adirondack region of New York state; Lake Whatchamacallit (surface area = 4 mi2, average depth = 42 ft). The pH of Lake Whatchamacallit has been measured to be 4.0 (an unfortunate result of acid deposition), which is just a little too acidic to support aquatic life; i.e. Lake Whatchamacallit is for all intents and purposes, dead.

Farmer Jones, whose property adjoins the lake, knows that when the soil on her farm is too acidic she adds lime (crushed limestone i.e. calcium carbonate, CaCO3) to it in order to increase the pH of the soil and make it suitable for planting. She gathers all of her neighbors for a meeting with the state's department of natural resources. She proposes that they lime the lake in order to increase its pH and then to restock the lake with new fish.

How many tons of lime would have to be added to Lake Whatchamacallit in order to raise the pH from 4.0 to 7.0? Use the following facts to help you determine your answer.

- 1 oz of lime will raise the pH of 5,700 liters of lake-water from 4.0 to 7.0

- the cost of lime is about 5.6 per pound

- when lime dissolves in water heat is given off, such that when 100g dissolves in water it gives off enough heat to increase the temperature of 3 liters of water by 1 degree C

- 1 mi = 5280 ft; 16 oz = 1 lb; 2000 lb = 1 ton; 1 ft3 lake-water = 28.3 liters

Hint: Start by finding the volume of the lake.

Discuss the implications of your results.

##### Solution Summary

This solution analyses lime content in a lake by looking at its surface area, depth and density as well as the implications of the pH findings.

##### Solution Preview

because,
surface area of lake (A) = 4 mi2 = 4*5280^2 = 1.115136*10^8 ft^2
depth of the lake (h) = 42 ft
hence, the volume of the lake:
V = A*h =1.115136*10^8*42 = 46835712*10^9 ft^3
because 1ft^3 = 28.3 lit
hence,
V = 46835712*10^9*28.3 = 1.3254506496*10^(11) lit
density of water = 1 kg/lit
because, to increase the pH value of 5700 lit of lake ...

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• MSc , Pune University, India
• PhD (IP), Pune University, India
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