Question

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants for sulfurous acid are K a1 = 1.4 × 10 − 2 and K a2 = 6.3 × 10 − 8 . Let's start by first calculating the concentration of the spectator ion.

Answer 1

Explanation:

Reaction equation is as follows.

      $Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

                                  = 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$.

As,     $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

                       = $\frac{10^{-14}}{6.3 \times 10^{-8}}$

                       = $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

          $SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        $K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

 $1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

        $x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

                x = 0.0003 M

So,   x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

                     = 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

              $HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              $K_{b} = \frac{x^{2}}{0.0003 - x}$

        $K_{b} = \frac{K_{w}}{K_{a_{1}}}$

                      = $\frac{10^{-14}}{1.4 \times 10^{-2}}$

                      = $7.14 \times 10^{-13}$

Therefore,  $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

                              = $0.00214 \times 10^{-13}$

                     x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

    $[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

                = $\frac{10^{-14}}{0.0003}$

                = $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.