in this module we\'ve been discussing fossil fuels and the impacts of their use.
ID: 914309 • Letter: I
Question
in this module we've been discussing fossil fuels and the impacts of their use. One of the impacts is an increase in acid deposition (or acid rain as many people refer to it). An acid can be neutralized by a base (also known as an alkali). Acid deposition can be neutralized by the addition of agricultural lime (an alkaline rock, not the tasty green fruit). The following exercise asks you to calculate how much agricultural lime it would take to increase the pH of a small lake. Remember that acids have lower pHs and bases have higher pHs, so that as I add more base, the pH will increase.
Consider a small lake in the Adirondack region of New York state; Lake Whatchamacallit (surface area = 4.18 mi2, average depth = 31.31 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 agricultural 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 much would it cost in dollars 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 agricultural lime is about $17 per ton
when lime dissolves in water heat is given off, such that when 100 g 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
Explanation / Answer
ANSWER:
CONCEPT: Determine the water in lake, and then determiine the amount of lime needed to riase the pH of this water.
Volume of Water in lake = volume of lake = surface area X average depth
Volume of Water in lake = (4.18 mi2 X 31.31ft)
convert 4.18 mi2 to ft2
4.18 mi2 = (4.18 X 5280) = 22070.4ft2
Volume of Water in lake =22070.4 X 31.31 = 691024.224ft3
Volume of Water in liters = 691024.224 X 28.3 = 19555985.53L
5700 L of water require 1 oz of lime to raise the pH,
Hence 1 L will require 1 / 5700 oz of lime
19555985.53L of lake water will require (1 / 5700) X 19555985.53 = 3430.87 oz of lime
convert 3430.87 oz to lb
3430.87oz = 3430.87 oz / 16 lb = 214.4 lb
convert 214.4lb to ton
191.69 lb = 214.4/ 2000 = 0.107 ton
Hence 0.107 ton of lime are required to raise the pH of lake from 4 to 7.
Cost of lime = 17 X 0.107 = 1.822 $
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