# Question 558b8

May 11, 2015

Sodium sulfate will dissociate completely in aqueous solution to give sodium cations, $N {a}^{+}$, and sulfite anions, $S {O}_{3}^{2 -}$.

$N {a}_{2} S {O}_{3 \left(s\right)} \to \textcolor{red}{2} N {a}_{\left(a q\right)}^{+} + S {O}_{3 \left(a q\right)}^{2 -}$

Notice that 1 mole of sodium sulfite will produce $\textcolor{red}{2}$ moles of sodium cations and 1 mole of sulfite anions. This means that you get

$\left[N {a}^{+}\right] = 2 \cdot \left[N {a}_{2} S {O}_{3}\right] = 2 \cdot 0.300 = \text{0.600 M}$

$\left[S {O}_{3}^{2 -}\right] = \left[N {a}_{2} S {O}_{3}\right] = \text{0.300 M}$

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

${K}_{b} = {K}_{W} / {K}_{a 2} = {10}^{- 14} / \left(6.3 \cdot {10}^{- 8}\right) = 1.59 \cdot {10}^{- 7}$

Use an ICE table for the equilibrium reaction that will be established to determine the concentration of the bisulfate and hydroxide ions

$\text{ } S {O}_{3 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H S {O}_{3 \left(a q\right)}^{-} + O {H}_{\left(a q\right)}^{-}$
I......0.300.................................0.....................0
C......(-x)....................................(+x).................(+x)
E...0.300-x................................x......................x

The base dissociation constant will be equal to

${K}_{b} = \frac{\left[H S {O}_{3}^{-}\right] \cdot \left[O {H}^{-}\right]}{\left[S {O}_{3}^{2 -}\right]} = \frac{x \cdot x}{0.300 - x} = {x}^{2} / \left(0.300 - x\right)$

Because the value of ${K}_{b}$ is so small, you can approximate (0.300 - x) with 0.300. This means that

${K}_{b} = {x}^{2} / 0.300 = 1.59 \cdot {10}^{- 7} \implies x = 2.18 \cdot {10}^{- 4}$

As a result, you'll get

$\left[O {H}^{-}\right] = 2.18 \cdot {10}^{- 4} \text{M}$

[HSO_3""^(-)] = 2.18 * 10^(-4)"M"

$\left[S {O}_{3}^{2 -}\right] = 0.300 - 2.18 \cdot {10}^{- 4} \cong \text{0.300 M}$

Now for the tricky part. The bisulfate ion can also act as a base and react with water to form sulfurous acid, ${H}_{2} S {O}_{3}$. The problem with sulfurous acid is that it doesn't exist in aqueous solution in that form, but rather as sulfur dioxid, $S {O}_{2}$, 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

${K}_{b} = {K}_{W} / {K}_{a 1} = {10}^{- 14} / \left(1.4 \cdot {10}^{- 2}\right) = 7.14 \cdot {10}^{- 13}$

When the bisulfate ion reacts with water, it'll form

$\text{ } H S {O}_{3 \left(a q\right)}^{-} + \cancel{{H}_{2} {O}_{\left(l\right)}} r i g h t \le f t h a r p \infty n s S {O}_{2 \left(a q\right)} + \cancel{{H}_{2} {O}_{\left(l\right)}} + O {H}_{\left(a q\right)}^{-}$
I....$2.18 \cdot {10}^{- 4}$.............................0...................................$2.18 \cdot {10}^{- 4}$
C.........(-x)......................................(+x)........................................(+x)
E...$2.18 \cdot {10}^{- 4} \text{-x}$.......................x..................................$2.18 \cdot {10}^{- 4} \text{+x}$

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 $\left(2.18 \cdot {10}^{- 4} \text{-x}\right)$ and $\left(2.18 \cdot {10}^{- 4} \text{+x}\right)$ with $2.18 \cdot {10}^{- 4}$. This will get you

${K}_{b} = \frac{\cancel{2.18 \cdot {10}^{- 4}} \cdot x}{\cancel{2.18 \cdot {10}^{- 4}}} = x = 7.14 \cdot {10}^{- 13}$

As a result, the concentrations of all the species listed will be

$\left[N {a}^{+}\right] = \textcolor{g r e e n}{\text{0.600 M}}$
$\left[S {O}_{3}^{2 -}\right] = \textcolor{g r e e n}{\text{0.300 M}}$
$\left[H S {O}_{3} \text{^(-)] = 2.18 * 10^(-4)-7.14 * 10^(-13) ~= color(green)(2.18 * 10^(-4)"M}\right)$
$\left[O {H}^{-}\right] = 2.18 \cdot {10}^{- 4} + 7.14 \cdot {10}^{- 13} \cong \textcolor{g r e e n}{2.18 \cdot {10}^{- 4} \text{M}}$

$\left[{H}^{+}\right] = {10}^{- 14} / \left(\left[O {H}^{-}\right]\right) = {10}^{- 14} / \left(2.18 \cdot {10}^{- 4}\right) = \textcolor{g r e e n}{4.59 \cdot {10}^{- 11} \text{M}}$