Calculate the equilibrium constant for the dissolution of calcium carbonate at pH=6 in the presence of EDTA as shown below: CaCO3(s) + EDTA^4- (aq) + H+(aq) --> [Ca(EDTA)]2-(aq) + HCO3 (-1 charge)(aq) K=?
Ksp=4.5 x 10^-9 for CaCO3; kf'=8.04 x 10^5 for [Ca(EDTA)]2- at pH=6 (kf' is the formation constant corrected for pH=6, in other words you do not need to calculate the fractional dissociation in this problem) pka1=6.35; ka1=4.46 x 10^-7 and pka2=10.33 (ka2=4.69 x 10^-11) for carbonic acid, H2CO3