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What is the number of moles of hydrochloric acid needed to neutralize 25.0 mL of 0.10 M NaOH?

A. 0.0025 moles
B. 0.005 moles
C. 0.010 moles
D. 0.025 moles

1 Answer

3 votes

Final answer:

The number of moles of hydrochloric acid required to neutralize 25.0 mL of 0.10 M NaOH is 0.0025 moles, as the reaction between HCl and NaOH is in a one-to-one mole ratio. The correct option is A.

Step-by-step explanation:

The question is about finding the number of moles of hydrochloric acid (HCl) needed to neutralize 25.0 mL of 0.10 M sodium hydroxide (NaOH). To solve this, we look at the balanced chemical equation for the reaction between HCl and NaOH, which shows a one-to-one mole ratio:

HCl + NaOH → NaCl + H2O

Using this ratio, we calculate the moles of NaOH present in the solution:

Number of moles of NaOH = Volume (L) × Molarity (M) = 0.025 L × 0.10 M = 0.0025 moles

Since it is a 1:1 reaction, the moles of HCl needed to neutralize the NaOH would be the same. Therefore, the answer is 0.0025 moles of HCl, which corresponds to option A. The correct option is A.

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