*If the solubility of CO2 in pure water*

*If the solubility of CO2 in pure water*

#### If the solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M, and if we assume that all of the dissolved CO2 is in the form of carbonic acid H2CO3, what is the pH of the solution of H2CO3?

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As H2CO3 is a diprotic acid, the two acid dissociation constants are Ka1 and Ka2 respectively. Here we can determine the pH by considering only Ka1, thus treating it as a monoprotic acid.

H2CO3 (aq) = H+ (aq) + HCO3 (aq)

For H2C03,

Initial = 0.0037M

Change = -x M

Equilibrium = (0.0037 – x) M

For H+,

Initial = 0

Change = +x M

Equilibrium = x M

For HCO3,

Initial = 0

Change = +x M

Equilibrium = x M

Ka1 = ( [H+] [HCO3] ) / [H2CO3]

= [ (x) (x) ] / 0.0037 – x

= 4.3 * 10^-7

x = 4.0 * 10^-5

Solving this, we get:

0.0037 – x = 0.0037

[ (x) (x) ] / 0.0037 = 4.3 * 10^-7

x^2 = ( 0.0037) ( 4.3 * 10^-7 )

= 1.6 * 10^-9

Thus, x = [H+] = [HCO3]

= √ 1.6 * 10^-9

= 4.0 * 10^-5 M