We’re being asked to calculate the pH of a 0.1 M aqueous solution of sodium sulfite (Na2SO3). Na2SO3 is an ionic compound it will break up into its ions in the solution:
Na2SO3(aq)→ 2 Na+(aq) + SO32-(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
• SO32- is the basic form of the diprotic acid H2SO3. A diprotic acid means it can donate two protons (H+) and it will have two equilibrium reactions.
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 Kb 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, SO2, acts as a diprotic acid in aqueous solution. At 25°C the acidity constants are:
SO2 (aq) + 2H 2O (l) ⇋ HSO3– (aq) + H3O+ (aq) K a1 = 1.54 x 10–2
HSO3– (aq) + H2O (l) ⇋ SO32– (aq) + H3O+ (aq) K a2 = 1.02 x 10–7
Calculate the pH of a 0.1 M aqueous solution of sodium sulfite (Na 2SO3).