Question

1. The pH of natural water and the effects of global warming

a) It is a well-known fact that the increase in atmospheric CO2 concentration will result in global warming. Such effects will likely impact the pH of natural water and thereby the aquatic ecosystem as well. Explain qualitively to your layman friends whether pH of natural water will be increased or decreased based on acid-base equilibrium?

b) The current (Year 2013) atmospheric CO2 concentration is 370 ppm. It is estimated that if we continue to exploit the world’s fossil fuel reserves at the same rate, the atmospheric CO2 concentration will reach 1,100 ppm in Year 2400 (Williams, 2001). Justify your answer in (1) quantitatively by calculating the pHs when CO2 concentrations are 370 ppm (Year 2006) and 1,100 ppm (Year 2400), respectively.

Assuming: Pure water is in equilibrium with CO2 at 25oC. Water vapor pressure is 0.031 atm. Henry’s Law constant 3.38x10-2 (mol/L) x atm-1 at 25oC for CO2. Ka1 = 4.45x10-7; Ka2 = 4.69x10-11.

Neglect all other processes such as complexation, and precipitation.

Answer 1

(A)

Carbonic acid (H₂CO₃), which dissociates to a bicarbonate ion and a proton (H⁺ ).

Acid/base metabolism in our0 body is regulated by the following chemical equation:

CO₂ + H₂O ↔ H₂CO₃ ↔ H⁺ + HCO₃^-

Formation of water (i.e. from H⁺ and OH^-) is an endothermic reaction. During global warming, water ionizes to give H⁺ and the pH decreses i.e. acidic characteristic of water increases.

(B)

Using Henry's law,

solubility of CO₂ = 3.38 x 10^-2 mol/L.atm x 0.031 atm

= 1.05 x 10^-3 M

H₂CO₃ <------> H⁺ + HCO₃^-

I 1.05 x 10^-3 0           0

E 1.05 x 10^-3 -x + x +x

Ka1 = [H⁺][HCO₃^-]/[H₂CO₃]

4.45 x 10^-7 = x² / 1.05 x 10^-3 -x

4.45 x 10^-7 (1.05 x 10^-3 -x) - x² = 0

4.67 x 10^-10 - 4.45 x 10^-7 x -x² = 0

x² + 4.45 x 10^-7 x - 4.67 x 10^-10 = 0

Using quadratic equation a = 1, b = 4.45 x 10^-7 ,c =  - 4.67 x 10^-10

     x =​ −b±√​b​2​​−4ac​​​​​ / 2a

x = {- 4.45 x 10^-7 ±√​ (4.45 x 10^-7)² - 4*1* - 4.67 x 10^-10 } / 2*1

x = {- 4.45 x 10^-7 ±√​19.8x 10^-14 + 18.7 x 10^-10 } / 2

x = {- 4.45 x 10^-7 ±√​1.87x 10^-9}/2

x = {- 4.45 x 10^-7 ± 0.0000432} / 2

x1 = {- 4.45 x 10^-7 + 0.0000432} / 2 = 0.0000214 M = x

x2 = {- 4.45 x 10^-7 - 0.0000432} / 2 = negative concentration, thus neglected.

Now,

x = [H⁺] = [HCO₃^-] = 0.0000214 M

HCO₃^- <-------> H⁺ + CO₃^-2

I 0.0000214 0.0000214 0

E 0.0000214 0.0000214 + x x

Ka2 = (0.0000214+x) (x) / (0.0000214-x)

4.69 x 10^-11 = (0.0000214+x) (x) / (0.0000214 -x )

x is very less compared to 0.0000214, so neglect x.

4.69 x 10^-11 = x

Now, [H⁺] = 0.0000214 M

pH = -log [H⁺]

= -log 0.0000214

pH = 4.67 (Acidic)