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 (CaCO3). How much limestone in kilograms is required to completely neutralize a 3.8 × 109 L lake with a pH of 5.5?

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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