Titration of 50.0mL of 0.130mol/L sulfurous acid with 0.14mol/L sodium hydroxide.
What would the first and second equivalence point be on the graph?
To determine the first and second equivalence points on the graph, we need to look at the reaction between sulfurous acid (H2SO3) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is:
H2SO3 + 2NaOH -> Na2SO3 + 2H2O
From the equation, we can see that for each mole of sulfurous acid, 2 moles of sodium hydroxide are required for complete neutralization.
Given:
- Volume of sulfurous acid (H2SO3) = 50.0 mL = 0.050 L
- Concentration of sulfurous acid (H2SO3) = 0.130 mol/L
- Concentration of sodium hydroxide (NaOH) = 0.14 mol/L
First, we need to determine the number of moles of sulfurous acid in the 50.0 mL solution:
Moles of H2SO3 = volume (L) x concentration (mol/L)
Moles of H2SO3 = 0.050 L x 0.130 mol/L
Moles of H2SO3 = 0.0065 mol
Since sulfurous acid reacts with sodium hydroxide in a 1:2 ratio, we would need twice as many moles of sodium hydroxide for neutralization. Therefore, the first equivalence point occurs when half of the moles of sulfurous acid are neutralized.
First equivalence point:
Moles of NaOH required = 0.5 x Moles of H2SO3
Moles of NaOH required = 0.5 x 0.0065 mol
Moles of NaOH required = 0.00325 mol
To determine the volume of sodium hydroxide required at the first equivalence point, we can use the equation:
Volume (L) = Moles / Concentration (mol/L)
Volume of NaOH required = 0.00325 mol / 0.14 mol/L
Volume of NaOH required ≈ 0.023 L or 23 mL
Therefore, the first equivalence point occurs at approximately 23 mL.
At the second equivalence point, all the moles of sulfurous acid have been neutralized. Therefore, the volume required for the second equivalence point is the sum of the volume required at the first equivalence point and the original volume of sulfurous acid.
Second equivalence point:
Volume of NaOH required = 0.050 L + 0.023 L
Volume of NaOH required = 0.073 L or 73 mL
Therefore, the second equivalence point occurs at approximately 73 mL on the graph.