Final answer:
To reach the third equivalence point in the titration of 17.5 mL of 0.135 M H₃PO₄, you would need to add 35.4 mL of 0.200 M KOH. The stoichiometry of the balanced equation helps in calculating the moles of KOH needed.
Step-by-step explanation:
To determine the volume of 0.200 M KOH needed to reach the third equivalence point in the titration of 17.5 mL of 0.135 M H₃PO₄, we need to use the stoichiometry of the balanced equation. The balanced equation for the reaction between H₃PO₄ and KOH is:
- H₃PO₄ (aq) + 3 KOH (aq) -> K₃PO₄ (aq) + 3 H₂O (l)
From the balanced equation, we can see that the stoichiometric ratio of H₃PO₄ to KOH is 1:3. Therefore, we need 3 times the moles of H₃PO₄ to neutralize it completely. First, calculate the moles of H₃PO₄ in 17.5 mL:
Moles of H₃PO₄ = concentration of H₃PO₄ x volume of H₃PO₄ = 0.135 M x 0.0175 L = 0.0023625 mol H₃PO₄
Since the stoichiometric ratio is 1:3, we would need 3 times the moles of H₃PO₄ of KOH:
Moles of KOH = 3 x moles of H₃PO₄ = 3 x 0.0023625 mol H₃PO₄ = 0.0070875 mol KOH
Now, using the molarity and the moles of KOH, we can calculate the volume:
Volume of KOH = moles of KOH / concentration of KOH = 0.0070875 mol / 0.200 M = 0.0354375 L = 35.4375 mL
Therefore, you would need to add 35.4 mL of 0.200 M KOH to reach the third equivalence point.