Question

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 1.8×10–4 bar in the atmosphere, what is the pH of water in equilibrium with the atmosphere? (For carbonic acid Ka1 = 4.46× 10–7 and Ka2 = 4.69× 10–11).

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation CO2(g) 근 CO2(aq) に0.032 The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 1.8x104 bar in the atmosphere, what is the pH of water in equilibrium with the atmosphere? (For carbonic acid Ka 4.46x 10-7 and Ka2 4.69x 10-"). Number 5.05 Incorrect. Start by using the equilibrium constant expression for the dissolution of gaseous CO2 to calculate [CO2, which equals [H2CO3l. CO,(aq CO

Answer 1

given partial pressure = 1.8*10^-4 bar

atmospheric pressure = 0.9869 bar

so moles fraction of CO2 in air =1.8*10^-4/0.9826 =1.8239*10^-4

lets do the caclculation for one mole of air so no of moles CO2 =1.8239*10^-4

CO2 [g] <--------------> CO2[aq]

c-x x

given K =0.032= x/[c-x] where c initial concentration of CO2

so x= Kc/[1+K]= 0.032*1.8239*10^-4/[1+0.032]=5.65*10^-6 M

this concentraion will be the initial CO2 concentration in water

as Ka1 = [concentration of H+ ion]^2/x =4.46*10^-7

[concentration of H+ ion]^2 =4.46*10^-7*5.65*10^-6=2.52*10^-12

gives [H+]=1.588*10^-6 M

so pH = -log 1.588*10^-6 =5.799