!! LONG ANSWER !!
Sodium sulfate will dissociate completely in aqueous solution to give sodium cations, Na^(+)Na+, and sulfite anions, SO_3^(2-)SO2−3.
Na_2SO_(3(s)) -> color(red)(2)Na_((aq))^(+) + SO_(3(aq))^(2-)Na2SO3(s)→2Na+(aq)+SO2−3(aq)
Notice that 1 mole of sodium sulfite will produce color(red)(2)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"[Na+]=2⋅[Na2SO3]=2⋅0.300=0.600 M
[SO_3^(2-)] = [Na_2SO_3] = "0.300 M"[SO2−3]=[Na2SO3]=0.300 M
The sulfite anion will act as a base and react with water to form the bisulfate ion, or HSO_3""^(-)HSO3−. The base dissociation constant, K_bKb, for the sulfite ion, will be equal to
K_b = K_W/K_(a2) = 10^(-14)/(6.3 * 10^(-8)) = 1.59 * 10^(-7)Kb=KWKa2=10−146.3⋅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))^(-) SO2−3(aq)+H2O(l)⇌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)Kb=[HSO−3]⋅[OH−][SO2−3]=x⋅x0.300−x=x20.300−x
Because the value of K_bKb 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)Kb=x20.300=1.59⋅10−7⇒x=2.18⋅10−4
As a result, you'll get
[OH^(-)] = 2.18 * 10^(-4)"M"[OH−]=2.18⋅10−4M
[HSO_3""^(-)] = 2.18 * 10^(-4)"M"[HSO3−]=2.18⋅10−4M
[SO_3^(2-)] = 0.300 - 2.18 * 10^(-4) ~= "0.300 M"[SO2−3]=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_3H2SO3. The problem with sulfurous acid is that it doesn't exist in aqueous solution in that form, but rather as sulfur dioxid, SO_2SO2, 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")
