!! LONG ANSWER !!
Sodium sulfate will dissociate completely in aqueous solution to give sodium cations, #Na^(+)#, and sulfite anions, #SO_3^(2-)#.
#Na_2SO_(3(s)) -> color(red)(2)Na_((aq))^(+) + SO_(3(aq))^(2-)#
Notice that 1 mole of sodium sulfite will produce #color(red)(2)# moles of sodium cations and 1 mole of sulfite anions. This means that you get
#[Na^(+)] = 2 * [Na_2SO_3] = 2 * 0.300 = "0.600 M"#
#[SO_3^(2-)] = [Na_2SO_3] = "0.300 M"#
The sulfite anion will act as a base and react with water to form the bisulfate ion, or #HSO_3""^(-)#. The base dissociation constant, #K_b#, for the sulfite ion, will be equal to
#K_b = K_W/K_(a2) = 10^(-14)/(6.3 * 10^(-8)) = 1.59 * 10^(-7)#
Use an ICE table for the equilibrium reaction that will be established to determine the concentration of the bisulfate and hydroxide ions
#" "SO_(3(aq))^(2-) + H_2O_((l)) rightleftharpoons HSO_(3(aq))^(-) + OH_((aq))^(-)#
I......0.300.................................0.....................0
C......(-x)....................................(+x).................(+x)
E...0.300-x................................x......................x
The base dissociation constant will be equal to
#K_b = ([HSO_3^(-)] * [OH^(-)])/([SO_3^(2-)]) = (x * x)/(0.300 - x) = x^2/(0.300 - x)#
Because the value of #K_b# is so small, you can approximate (0.300 - x) with 0.300. This means that
#K_b = x^2/0.300 = 1.59 * 10^(-7) => x = 2.18 * 10^(-4)#
As a result, you'll get
#[OH^(-)] = 2.18 * 10^(-4)"M"#
#[HSO_3""^(-)] = 2.18 * 10^(-4)"M"#
#[SO_3^(2-)] = 0.300 - 2.18 * 10^(-4) ~= "0.300 M"#
Now for the tricky part. The bisulfate ion can also act as a base and react with water to form sulfurous acid, #H_2SO_3#. The problem with sulfurous acid is that it doesn't exist in aqueous solution in that form, but rather as sulfur dioxid, #SO_2#, and water.
#underbrace(SO_2 + H_2O)_("color(blue)(H_2SO_3)) + H_2O_((l)) rightleftharpoons HSO_(3(aq))^(-) + H_3O_((aq))^(+)#
The base dissociation constant for the bisulfate ion will be
#K_b = K_W/K_(a1) = 10^(-14)/(1.4 * 10^(-2)) = 7.14 * 10^(-13)#
When the bisulfate ion reacts with water, it'll form
#" "HSO_(3(aq))^(-) + cancel(H_2O_((l))) rightleftharpoons SO_(2(aq)) + cancel(H_2O_((l))) + OH_((aq))^(-)#
I....#2.18 * 10^(-4)#.............................0...................................#2.18 * 10^(-4)#
C.........(-x)......................................(+x)........................................(+x)
E...#2.18 * 10^(-4)"-x"#.......................x..................................#2.18 * 10^(-4)"+x"#
#K_b = ([SO_2] * [OH^(-)])/([HSO_3""^(-)]) = ((2.18 * 10^(-4) + x) * x)/(2.18 * 10^(-4)-x)#
Once again, the very, very small value of #K_b# will allow you to approximate #(2.18 * 10^(-4)"-x")# and #(2.18 * 10^(-4)"+x")# with #2.18 * 10^(-4)#. This will get you
#K_b = (cancel(2.18 * 10^(-4)) * x)/(cancel(2.18 * 10^(-4))) = x = 7.14 * 10^(-13)#
As a result, the concentrations of all the species listed will be
#[Na^(+)] = color(green)("0.600 M")#
#[SO_3^(2-)] = color(green)("0.300 M")#
#[HSO_3""^(-)] = 2.18 * 10^(-4)-7.14 * 10^(-13) ~= color(green)(2.18 * 10^(-4)"M")#
#[OH^(-)] = 2.18 * 10^(-4) + 7.14 * 10^(-13) ~= color(green)(2.18 * 10^(-4)"M")#
#[H^(+)] = 10^(-14)/([OH^(-)]) = 10^(-14)/(2.18 * 10^(-4)) = color(green)(4.59 * 10^(-11)"M")#
