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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^+]

Chemistry Homework Help D

!! 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 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”)##







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