3 The mean monthly pH values of rainfall in Guiyang city in

3. The mean monthly pH values of rainfall in Guiyang city in Guizhou province in southern China in 1984 were as follows: Jan 3.9, 4.0, 3.8, 4.1, 4.0, 4.5, 4.5, 4.1, 3.7, 3.8, 3.7, 3.4 Dec For the same year, the measurements at Luizhang, an adjacent rural area were: Jan 4.3, 4.4, 4.4, 4.2, 4.5, 4.9, 4.9, 4.6, 4.8,4.3, 5.4, 5.4 Dec (a) Calculate the mean monthly pH of the rain at the two locations. (b) What is the ratio of the mean hydrogen ion activity at the two sites? (c) What is likely to be the most important anion in the rainfall at the urban site? Refer to Dianwu, Z. and X. Jiling, Acidification in southwestern China, In Acidification in tropical countries (ed. Rodhe, H. and R. Herrera), John Wiley and Sons, Chichester; 1988

Solution

Solution of a) mean pH of rain in city in a year = (sum of ph of monthly rain )/ (totoal no. of months)

mean pH of rain in city in a year= (3.9+4.0+3.8+4.1+4.0+4.5+4.5+4.1+3.7+3.8+3.7+3.4)/12

= 47.5/12

= 3.958

mean pH of rain in rural area in a year= (56.1)/12

= 4.675

Solution of b) we know the formula a(H) = K(a) x c(H)

where: a(H) = Hydrogen activity in the solution
K(a) = Hydrogen activity coefficient
c(H) = Hydrogen concentration in the solution

let us say for city a1, c1, k1

for rural area a2, c2, k2

as we can understand that the solvent water is same in both the place so k1 and k2 should be same i.e. k1= k2

so a1/ a2 = c1(H) / c2(H)

we know pH=log10[cH]

so we can rearrange this formula as c(H) =10^-pH

so we calculate c(H) using the average value of pH separately for both the sites

for city c1(H) = 10^-3.958 = 1.10154* 10^-4

for rural area c2(H) = 10^-4.675 = 2.11349* 10^-4

now using: a1/ a2 = c1(H) / c2(H)

=(1.10154* 10^-4) / (2.11349* 10^-4)

= 5.2119 this is the ratio of mean hydrogen activity

Solution c) In Urban site the average pH is 3.958 . the acids present in rain are Sulphuric acid , Nitric acid.

Anion in rainfall will be Sulphate anion (SO₄^2-) and nitrate anion (NO₃^-)