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1 Answers

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