We’re being asked to calculate the pH of a 0.1 M aqueous solution of sodium sulfite (Na_{2}SO_{3}). Na_{2}SO_{3} is an ionic compound it will break up into its ions in the solution:

**Na _{2}SO_{3(aq)}**

• Main group metals should have a +3 charge or higher to be considered acidic. Na^{+} has only a +1 charge and is a neutral ion so it will not contribute to the pH of the solution

• SO_{3}^{2-} is the **basic **form of the diprotic acid H_{2}SO_{3}. A diprotic acid means it can donate two protons (H^{+}) and it will have two equilibrium reactions.

${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{3}}\mathbf{}\underset{{\mathbf{K}}_{{\mathbf{b}}_{\mathbf{2}}}}{\overset{{\mathbf{K}}_{{\mathbf{a}}_{\mathbf{1}}}}{\mathbf{\rightleftharpoons}}}\mathbf{}{{{\mathbf{HSO}}}_{{\mathbf{3}}}}^{{\mathbf{-}}}{\mathbf{}}\underset{{\mathbf{K}}_{{\mathbf{b}}_{\mathbf{1}}}}{\overset{{{\mathbf{K}}_{\mathbf{a}}}_{\mathbf{2}}}{\mathbf{\rightleftharpoons}}}{\mathbf{}}{{{\mathbf{SO}}}_{{\mathbf{3}}}}^{\mathbf{2}\mathbf{-}}$

**We're going to use the following steps to calculate for the pH of the solution: **

Step 1: Construct an ICE chart for the equilibrium reaction.

Step 2: Write the K_{b} expression.

Step 3: Calculate for the equilibrium concentration.

Step 4: Calculate pOH.

Step 5: Calculate pH.

In the atmosphere, sulfur dioxide and nitrogen dioxide react with water to give sulfurous acid and nitric acid. This produces “acid rain” which has a pH below 7 and can dissolve limestone.

Sulfur dioxide, SO_{2}, acts as a diprotic acid in aqueous solution. At 25°C the acidity constants are:

SO_{2} (aq) + 2H _{2}O (l) ⇋ HSO_{3}^{–} (aq) + H_{3}O^{+ }(aq) K _{a1} = 1.54 x 10^{–2}

HSO_{3}^{– }(aq) + H_{2}O (l) ⇋ SO_{3}^{2– }(aq) + H_{3}O^{+ }(aq) K _{a2} = 1.02 x 10^{–7}

Calculate the pH of a 0.1 M aqueous solution of sodium sulfite (Na _{2}SO_{3}).

A. 4.00

B. 10.00

C. 7.00

D. 1.00

E. 6.59