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QUESTION

Calculate the concentrations of all species in a 0.300 M Na_2SO_3 (sodium sulfite) solution? The ionization constants for sulfurous acid are K_(a1) = 1.4* 10^(–2) and K_(a2) = 6.3 * 10^(–8). [Na^+] [SO_3^-2] [HSO_3^-] [OH^-] [H^+]

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 of the bisulfate and hydroxide ions

" "SO_(3(aq))^(2-) + H_2O_((l)) rightleftharpoons HSO_(3(aq))^(-) + OH_((aq))^(-)I......0.300.................................0.....................0C......(-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 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")