Question

Write the equation for the reaction of HSO3- with water in which it acts like an ACID and identify the acid, base, conjugate acid, and conjugate base. Identify the strongest acid, then write an expression for Ka. The value of Ka at 35 degrees celsius is 1.23 x 10^-7. Draw an arrow indicating which side is favored by equilibrium.

Answer 1

Bronsted Lowery Acid/Bases

First, let us define Bronsted Lowry acid/base:

Bronsted Lowry acid: any species that will donate H+ (protons) in solution, and makes pH lower (i.e HCl)

Bronsted Lowry base: any species that will accept H+ (protons) in solution, and makes pH higher (NH3 will accept H+ to form NH4+)

Typically, acid/bases are shown in the left (reactants)

when we write the products:

Bronsted Lowery conjugate base = the base formed when the B.L. acid donates its H+ proton ( i.e. HCl -> Cl-

Bronsted Lowery conjugate acid = the acid formed when the B.L. base accept its H+ proton ( i.e. NH4+ has accept H+ proton)

Note that, typically conjugate bases/acids are shown in the right (product) side

So, from your reaction:

HSO3-(aq) + H2O(l) <> SO3-2(aq) + H3O(l)

Left side either acid/base, right base are always conjugate (acids/bases)

then

HSO3- is actic as an ACID, so it is the B.L. acid

H2O acts as the base, so ti is the B-L. Base

SO3-2 is the conjugate base of HSO3-, since it is donating H+, it can accetp H+ as a conjguate

H3O+ is the cnojugate acid, since it can donate H+ later

at 35ºC:

pKa for HSO3- = 6.5 approx

pKa for H2O = 35

therefore, the stronger acid is HSO3-

that is, HSO3- will donate H+ and H"O will most likely accept it

the arrow goes as ---->