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) +...
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