(OH- is a base)
(H+ is an acid)
Therefore by adding water to HSO3, the OH- ion is produced therefore it is an Arrhenius base.
Interesting and very difficult to put into x amount of letters.
H2SO3 will ionize to H+ + HSO3-. The Keq = [H+][HSO3-]/[H2SO3]HSO3- will ionize to H+ + SO3^2-. The Keq = [H+][SO^2-]/[HSO3-] ... very small value
HSO3-
This is not one formula, but a complete set of the chemical reaction equations:SO2 + H2O [H2SO3] HSO3− + H+ (Sulfurous acid)Ka = 1.54×10−2; pKa = 1.81.
There are two consecutive equilibrium reactions:Hydration SO2,g + H2O H2SO3,aqProtolysis H2SO3,aq + H2O HSO3-aq + H3O+aq
The conjugate base and conjugate acid for HS04 is: Conjugate acid is H2SO4 Conjugate base is SO42
SO32
Fe(II)(HSO3)2 Iron(II) bisulphite . NB sulphurous acid is H2SO3 . Its anion is Sulphite (SO3^2-) The ''bi' means the sulphurous acid has only 'lost' ONE(1) hydrogen Hence the sulphite becomes the 'bisulphite' (HSO3^-) Since the iron is in oxidation state '2' , its cation is Fe^2+. Hence it needs two 'bisulphite' anions to balance the charges. NNB Do not confuse the 'bi' to mean 'two'. In this case it refers to the loss of hydrogen from the acid, in a similar way to sodium bi-carbonate (NaHCO3). Hope that helps!!!!
H2so4 + h2o <==> hso3^-1 + h3o+
The chemical formula of the Marshall's acid is H2S2O8 HSO3-O-O-SO3H.
Sulfur dioxide reacts with water to form sulfurous acid (SO2 + H2O --> HSO3)
Assuming you mean soluble in water - in theory, yes, it would be referred to as sulfurous acid - HOWEVER, there is no evidence that it actually exists in solution. The closest you would come to find is HSO3- and H3O+ from the H2SO3 dissociating into SO2 and 2 H2O as soon as it got into the water. See http://en.wikipedia.org/wiki/Sulfurous_acid and the references contained in that article