Question

How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H₃PO₄? H₃PO₄(aq)+2KOH(aq)⟶K₂HPO₄(aq)+2H₂O(l)

a) 15.40 mL
b) 30.80 mL
c) 46.20 mL
d) 61.60 mL

Answer 1

To find the required volume of a 0.1500-M KOH solution needed to titrate a 40.00 mL, 0.0656-M solution of H3PO4, we calculate the moles of H3PO4, use the stoichiometry of the reaction to find the moles of KOH needed, and then calculate the volume of KOH solution required, which should be 34.987 mL.

Step-by-step explanation:

The question involves a titration calculation in which a solution of KOH is used to titrate a solution of H3PO4. According to the balanced chemical equation provided, it takes 2 moles of KOH to neutralize 1 mole of H3PO4:

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

To solve for the required volume of KOH, we can use the molarity (M) and volume of H3PO4, and the molarity of KOH.

First, calculate the number of moles of H3PO4:

• Moles of H3PO4 = 0.0656 M × 0.04000 L = 0.002624 moles H3PO4

Then, use the stoichiometry from the equation to find the moles of KOH needed:

• Moles of KOH needed = 2 × (moles of H3PO4) = 2 × 0.002624 moles = 0.005248 moles KOH

Finally, calculate the volume of 0.1500 M KOH solution needed:

• Volume of KOH = (moles of KOH) / (molarity of KOH) = 0.005248 moles / 0.1500 M = 0.034987 L

Converting liters to milliliters (1 L = 1000 mL):

• Volume of KOH in mL = 0.034987 L × 1000 mL/L = 34.987 mL

This value does not match any of the given multiple-choice options, which suggests a possible mistake in those options or in the calculation.