Question

If the sample of H₃PO₄(aq) had a volume of 32.00 ml and the titration required 23.56 ml of 1.287 m KOH to reach the end point, what was the molarity of the H₃PO₄ solution?

Answer 1

The molarity of the H₃PO₄ solution is calculated by using the reaction's stoichiometry and the volumes and concentrations of the reacting solutions, resulting in a molarity of 0.5639 M.

Step-by-step explanation:

The student's question is regarding the calculation of the molarity of an H₃PO₄ solution after performing a titration with a KOH solution. Given the volume and molarity of the KOH solution and the volume of the H₃PO₄ solution, the molarity of the H₃PO₄ can be found using the formula:

M1V1 = M2V2

Where M1 and V1 are the molarity and volume of the KOH solution, and M2 and V2 are the molarity and volume of the H₃PO₄ solution, respectively.

Given:
V1 = 23.56 mL of KOH
M1 = 1.287 M KOH
V2 = 32.00 mL of H₃PO₄
M2 = ? (What we want to find)

First, convert the volumes from mL to L:

V1 = 23.56 mL × (1 L/1000 mL) = 0.02356 L
V2 = 32.00 mL × (1 L/1000 mL) = 0.03200 L

Using the balanced chemical equation for the titration:

H₃PO₄ (aq) + 2KOH (aq) → K₂HPO₄ (aq) + 2H₂O (1)

We can see that the reaction ratio of H₃PO₄ to KOH is 1:2. Therefore, we must account for this in our calculations:

M2 = (M1V1) / (2 × V2)

M2 = (1.287 M × 0.02356 L) / (2 × 0.03200 L)

M2 = 0.5639 M

Hence, the molarity of the H₃PO₄ solution is 0.5639 M.