Given the following equilibrium constants:
(1) `CaCO_(3)(s)toCa^(2+)(aq)+CO_(3)^(2-)(aq)K_(1)=10^(-8.4)`
(2) `HCO_(3)^(-)(aq)toH^(+)(aq)+CO_(3)^(2-)(aq)K_(2)=10^(-10.3)`
Calculate the value of K for the reaction
`CaCO_(3)(s)+H^(+)(aq)=Ca^(2+)(aq)+HCO_(3)^(-)(aq)`

Explanation

The net reaction is the sum of reaction 1 and the reverse of reaction 2:
`CaCO_(3)(s)toCa^(2+)(aq)+CO_(3)^(2-)(aq)K_(1)=10^(-8.4)`
`H^(+)(aq)+CO_(3)^(2-)(aq)toHCO_(3)^(-)(aq)K_(-2)=10^(-(-10.3))`
`bar(CaCO_(3)(s)+H^(+)(aq)toCa^(2+)(aq)+HCO_(3)^(-)(aq)K=K_(1)//K_(2)=10^((-8.4+10.3))=10^(+1.9))`
Comment:
This net reaction describes the dissolution of limestone by acid, it is responsible for the eroding effect of acid rain on buildings and statues. This is an example of a reaction that has practically no tendency to take place by itself (the dissolution of calcium carbonate) begin “driven” by a second reaction having a large equilibrium constant. From the standpoint of the LeChâtelier principle, the first reaction is “pulled to the right” by the removal of carbonate by the hydrogen ion. “Coupled” reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked in this way.