Solution for problem 118P Chapter 14
Lakes that have been acidified by acid rain (see Chemistry in the Environment box in
Introductory Chemistry | 5th Edition
Lakes that have been acidified by acid rain (see Chemistry in the Environment box in Section 5.9) can be neutralized by liming, the addition of limestone \(\mathrm{(CaCO_3)}\). How much limestone in kilograms is required to completely neutralize a \(\mathrm{3.8 \times 10^9~ L}\) lake with a pH of 5.5?
Equation Transcription:
Text Transcription:
(CaCO_3)
3.8 x 10^9 L
Solution: Here, we are going to calculate the mass of limestone required to completely neutralize the lake.
Step1:
pH is the negative logarithm to base 10 of hydrogen ion concentration.
pH = - log[H+ ]
Step2:
Given, pH of the Lake = 5.5
pH = - log[H+]
Therefore, [H+] = 10-pH
= 10-5.5
= 3.16 x 10-6 M
Therefore, number of moles of hydrogen ion = Volume of the lake x [H+]
= (3.8 x 109 L) x (3.16 x 10-6 mol/L)
= 12.0 x 103 mol
Step3:
The neutralization reaction taking place in the lake is:
CaCO3(s) + 2H+(aq) → Ca2+ + H2O + CO2
From the reaction, it is clear that for neutralization of 2 mol of hydrogen ion, 1 mole of CaCO3 is required.
Therefore, to neutralize 12 x 103 mol of hydrogen ion, number of moles of CaCO3 required = ½ x (1.2 x 103) = 6.0 x 103 mol.
Step4:
1 mol CaCO3 = molar mass of CaCO3
Therefore, 6.0 x 103 mol CaCO3 = 6.0 x 103 x molar mass of CaCO3
= 6.0 x 103 x 100.0869 g/mol
= 600.52 x 103 g
= 600.52 kg [1000 g = 1 kg]
Thus, 600.52 kg of limestone will be required.
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Chapter 14, Problem 118P is Solved
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Lakes that have been acidified by acid rain (see Chemistry in the Environment box in