Question

How many moles of diprotic acid are needed to neutralize 145 mL of 0.1 M NaOH?

A) 0.00725 moles
B) 0.0145 moles
C) 0.0725 moles
D) 0.145 moles

Answer 1

To neutralize 145 mL of 0.1 M NaOH, 0.00725 moles of diprotic acid are needed, because one mole of a diprotic acid can neutralize two moles of NaOH.

Step-by-step explanation:

To determine how many moles of diprotic acid are needed to neutralize 145 mL of 0.1 M NaOH, we first find the moles of NaOH present in the solution:

Number of moles of NaOH = Volume (L) × Molarity (M).

Volume of NaOH = 145 mL = 0.145 L (since 1000 mL = 1 L),
so:

Number of moles of NaOH = 0.145 L × 0.1 M = 0.0145 moles.

Since a diprotic acid can donate two protons (H+ ions) per molecule, one mole of diprotic acid can neutralize two moles of NaOH. To find the number of moles needed to neutralize the NaOH solution, we divide the moles of NaOH by two:

Number of moles of diprotic acid required = 0.0145 moles NaOH ÷ 2 = 0.00725 moles.

Therefore, the correct answer is A) 0.00725 moles.