122 BEYOND THE BACK OF THE ENVELOPE 9 is substituted into Eq. 7, a nonlinear equ

Question

122 BEYOND THE BACK OF THE ENVELOPE 9 is substituted into Eq. 7, a nonlinear equation The equation could, in principle, be solved eaton fot ( before, we will look for an approximate analytical solutacly, but me that the 5 × 10'"M of sulfur are en that H,SO, is completely dissociated to HSO;H completely dissociated to SOH. Then for eve three types of sulfate added, there are two mole of all moles of H' in solution or H1-2x 10 10 (10) Is this consistent with Eqs. 7 and 9? If Eq 10 is substituted Eq. 9, we calculate that H,Sol -10 110 11 not ather very mu a pt of 5.6. V iged in the cal EXERCISE 2 10-13 4 If Eqs. 10 and 11 are substituted into Eq. 7, 10' 10-134 101.4 10-134 10-14 . (12) od for every 2n So, indeed, Eq. 10 is correct and the pH of the precipitationis EXERCISE ted globally w fomly in glob pH=-logo [H+)-4. (13) This value is actually a good round-number estimate of the average of precipitation in regions of the world (such as northern Europe and the northeastern United States) where acid precipitation is a se s problem. You may wonder what happens to the rest of the SO. me of the SO, forms sulfate that is deposited in rain or snow in Canada, on the Atlantic Ocean, or in the southeastern United States. Much of the rest falls to earth in the form of dry particulate matter in a process called "dry deposition." This dry sulfate can form sulfuric articles fon be? pH EXERC niou So the last para way of a sh one of the acid when rain water eventually dissolves it. In an elegant pair ot Oppenheimer (1983a, 1983b) has applied box-model concepts to the analysis of regional acid precipitation formation a midwestern United States will result in a p H deposition in rain and snow in that area. way, it is illuminating go nd dispersal. His analysis shows that a reduction in SO, emissions in the eastern and onal reduction in Now that we have worked out the answer the long and rigorous g to go back and look at a shortcut. In paragraph Iscussion, above, we estimated the molar concentration tion to be 5 x 10-5 moles(Syliter. When H,SO, dis-

Explanation / Answer

CO2(g) --> CO2(aq) , CO2 gas goes into aqueous solution

CO2(aq) + H2O(l) ---> H2CO3(aq) CO2 forms carbonic acid with water

H2CO3(aq) --> H+ + HCO3-(aq) ; there is proton donation, i.e. acidification

Ka = [HCO3-][H+]/([H2CO3])

givne H2CO3 is too small, then

Ka = 10^-6.4

[H+] = 10^-4 + x; we account pH = 4 and the dissociation, x of the acid

10^-6.4 = (x)(10^-4+x) / (M-x)

assume M-X = M since x will be too small

(10^-6.4)(M) = x^2 + (10^-4)x

x^2 + (10^-4)x - (10^-6.4)(M) = 0

assume M = 0.0001 since CO2 is too small

x^2 + (10^-4)x - (10^-6.4)(0.0001) = 0

x^2 + (10^-4)x -3.9*10^-10 = 0

x = 0.00010003

[H+] = 10^-4 + 0.00010003 = 0.00020003

pH = -log(0.00020003 = 3.90

pH change is too little

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