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What volume (in mL) of 0.2550 M HCl is required to neutralize 50.00 mL of 0.8000 M NaOH?

User Fuad Kamal
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Final answer:

To neutralize 50.00 mL of 0.8000 M NaOH, you will need about 157 mL of 0.2550 M HCl. This is determined using stoichiometry and the concept of molarity in a neutralization reaction.

Step-by-step explanation:

The student's question asks how much volume (in mL) of a 0.2550 M solution of hydrochloric acid (HCl) is needed to neutralize 50.00 mL of a 0.8000 M solution of sodium hydroxide (NaOH). We can solve this problem using the concept of molarity and stoichiometry in a neutralization reaction

.

In a neutralization reaction, the stoichiometric ratio between HCl and NaOH is 1:1. This means for every mol of NaOH, one mol of HCl is required.

First, calculate moles of NaOH using the formula: moles = Molarity x Volume (in liters).

Then, use the stoichiometry to calculate the required moles of HCl. Because the stoichiometric ratio is 1:1, the moles of HCl required will be equal to the moles of NaOH.

Then, use the molarity of HCl to calculate the volume required using the formula: Volume = moles ÷ Molarity.

So the calculation will follow as:

  • Moles of NaOH = 0.8000 M x 0.05000 L = 0.04 mol
  • Moles of HCl required = 0.04 mol (since the stoichiometric ratio is 1:1)
  • Volume of HCl required = 0.04 mol ÷ 0.2550 M = 0.157 L or 157 mL.

Learn more about Neutralization Reaction

User Mike Shauneu
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