Introductory Chemistry - 5 Edition - Chapter 14 - Problem 118p
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# Lakes that have been acidified by acid rain (see Chemistry in the Environment box in

Introductory Chemistry | 5th Edition

Problem 118P

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

Accepted Solution
Step-by-Step Solution:
Step 1 of 3

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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