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lecture notes7

This is done by comparing the equilibrium constants

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Unformatted text preview: comparing the equilibrium constants for these two reactions. º H3O+ + CO32- HCO3- + H2O Ka HCO3- + H2O Kb º H2CO3 + OH- Notice that Ka is the same as Ka2 for H2CO3 and equals 4.69 x 10-11. Kb must be calculated from Ka for the conjugate acid (H2CO3) [Kb = 10-14/4.45 x 10-7 = 2.25 x 10-8]. Since Kb >> Ka for HCO3-, the base hydrolysis process is more important and we can treat this problem as a weak base in solution....
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