Solution Found!
Lakes that have been acidified by acid rain (see Chemistry in the Environment box in
Chapter 14, Problem 118P(choose chapter or problem)
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
Questions & Answers
QUESTION:
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
ANSWER:
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+]