When H2SO4, sulfuric acid, is added to water, it will ionize into

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The short answer is that the sulfuric acid will ionize to produce `H^+` (really `H_3O^+` ) and a mixture of `HSO_4^-` and `SO_4^(-2)` . No un-ionized `H_2SO_4` will be present.

Sulfuric acid is a diprotic acid, because it can potentially donate two hydrogen atoms (protons). Loss of the first proton,...

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The short answer is that the sulfuric acid will ionize to produce `H^+` (really `H_3O^+` ) and a mixture of `HSO_4^-` and `SO_4^(-2)` . No un-ionized `H_2SO_4` will be present.

Sulfuric acid is a diprotic acid, because it can potentially donate two hydrogen atoms (protons). Loss of the first proton, leaving `HSO_4^(-)` , is, however, different from loss of the second proton, leaving `SO_4^(2-)` . Sulfuric acid is considered a strong acid that will ionize completely in water, but this only applies to the first proton.

The equation for the first ionization, producing hydrogen sulfate ion, is

`H_2SO_4(aq) + H_2O(l) -gt H_3O^+(aq) + HSO_4^-` (aq)

and the equation for the second ionization, that of hydrogen sulfate ion producing sulfate ion, is

`HSO_4^(-)(aq) + H_2O(l) =H_3O^(+)(aq) + SO_4^(2-)(aq)`

This second ionization is weak and proceeds to equilibrium, with a `K_a` `= 1.0 x 10^(-2)` .

Thus, while all of the `H_2SO_4` is used up (ionized) in the first reaction, only some of the `HSO_4^(-)` ion is used up in the second reaction, so both and `SO_4^(-2)` will be present in the final mixture, along with `H_3O^+` .

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