You own a pond on some acreage near amajor metropolitan area. The pollution of t

Question

You own a pond on some acreage near amajor metropolitan area. The pollution of the city results in thedissolution

of gaseous oxides of nitrogen and sulfurin the rainwater. These oxides are the result of natural processesand the

combustion of fossil fuels (gas). Theresulting rainwater is now an acidic solution (acid rain). The acidrain has

caused your pond to become acidic (pH of4.30) and most of the aquatic life has died. You can reduce theacidity of

the pond by adding quicklime (CaO) tothe pond until the pH is 5.30, thus retuning the pond back tobeing

biologically active.

Make some simple assumptions and performthe necessary calculations to answer the question:How much

quicklime (CaO) would berequired to return a pond back to biologicalviability?

Some hints and things to thinkabout:

We can represent the acid in the pond asH3O+. The products of thereaction of quicklime and acid are water

and the calcium ion,Ca2+.

The pH is a measure of a solutionsacidity and can be calculated according to the equation: pH =–log[H3O+].

Here the symbol[H3O+] indicates the molarconcentration of acid. Molar concentration units aremoles/liter.

Use the pH values above and theinformation on pH to determine how many moles of quicklime areneeded.

There are still some things you need toassume to complete this calculation.

Explanation / Answer

Given that the pH of the rain water changesfrom 4.30 to 5.30.           Quick lime reacts with the acid in the fallowing way.               CaO + 2H3O+ ----> Ca2+ + 3H2O          Initial :   pH = 4.30                       - log[H3O+] = 4.30                                  [H3O+]= 5.01 x 10-5 M         Final :    pH = 5.30                                               - log[H3O+] =5.30                                  [H3O+]= 5.01 x 10-6 M         The  concentration ofacid reacted with lime =    5.01 x10-5 M - 5.01 x 10-6 M                                                                               = 4.509 x 10-5 M            If the pond has one litre of the rain waterthen   4.509 x 10-5 moles of theacid reacted with quick lime.          we have, CaO + 2H3O+ ----> Ca2+ +3H2O                       Number of moles of quick lime required to reactwith acid =  (4.509 x 10-5 / 2 )                                                                                                                  = 2.25 x 10-5 mol                  Amount of quick lime required = 2.25 x10-5 mol * 56.077 g/mol                                                                        = 0.00126 g                     

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