Question

3. Sulfite is a compound frequently added to preserve food and wine. In addition to inhibiting bacteria, it also prevents the oxidation of ethanol into acetic acid! a. It was previously discussed that sulfate, SO4 2– , can be categorized as a negligible base. Sulfite, SO3 2– , however, is basic. Considering the two structures, why is sulfite significantly more basic? b. Write the chemical equation and the equilibrium constant expression for the reaction between SO3 2– and water. c. Sulfurous acid, H2SO3 has a Ka1 = 1.23 x 10–2 and a Ka2 = 6.60 x 10–8 at 25 ºC. What is the pH of a 0.50M solution of Na2SO3?

Answer 1

Answer:

a)  The detailed explanation for the question being asked in section a is shown below

b) $SO_3^{2-}_{(aq)} \ \ \ + H_2O_{l} ------> HSO_3^-_{(aq)} + OH^-$

    Equilibrium constant expression  $K_c = \frac{[HSO_3^-][OH^-]}{]SO_3^{2-}]}$

c) pH = 10.45

Explanation:

a.

If we look at $SO_4^{2-}$, we will realize that it is a conjugate base of a strong acid $(H_2SO_4)$. However, the more stronger the acid, the weaker its conjugate base.

On the other hand $SO_3^{2-}$ is a conjugate base of a weak acid $H_2SO_3$ and the more weaker the acid, the stronger the basicity of its conjugate base.

b. The chemical equation for the reaction between $SO_3^{2-}$  and water can be expressed as follows:

$SO_3^{2-}_{(aq)} \ \ \ + H_2O_{l} ------> HSO_3^-_{(aq)} + OH^-$

Equilibrium constant $K_c = \frac{[HSO_3^-][OH^-]}{]SO_3^{2-}]}$

c.

The ICE Table is then constructed as follows:

                               

                             $SO_3^{2-}_{(aq)} \ \ \ + H_2O_{l} ------> HSO_3^-_{(aq)} + OH^-$

Initial  (M)              0.500                                                 0                 0

Change  (M)           - x                                                      + x               + x

Equilibrium  (M)   (0.500 - x)                                             x                x

$K_c = \frac{[HSO_3^-][OH^-]}{]SO_3^{2-}]}$

$K_c = \frac{K \omega }{Ka_2}$

where $K \omega$ is the ionic product of water = $10^{-14}$

$K_c = \frac{10^{-14}}{6.60*10^{-8}}$

$K_c = 1.52*10^{-7}$

However;

$1.52*10^{-7} = \frac{[x][x]}{[0.500 - x]}$

$1.52*10^{-7} = \frac{[x]^2}{[0.500]}$       since $K_c$ value is so small; then (0500 -x ) ≅ 0.500

$x^2 = 1.52*10^{-7}*0.5$

$x^2 = 7.6*10^{-8}$

$x = \sqrt{ 1.52*10^{-7}*0.5$

$x = 2.7568*10^{-4}$

$x = 0.00028$

$[OH^-] = x = 0.00028$

$pOH = -log \ [OH^-]$

$pOH = - log \ [0.00028]$

$pOH = 3.55$

pH + pOH = 14

pH + 3.55 = 14

pH = 14 - 3.55

pH = 10.45