Question

How many milliliters of 0.250M NaOH are required to neutralize 30.4 mL of 0.146M HCl?

Answer 1

27.2 milliliters of 0.250M NaOH are required to neutralize 30.4 mL of 0.146M HCl.

Step-by-step explanation:

To determine the volume of NaOH needed to neutralize HCl, we can use the equation
`\( \text{moles} = \text{Molarity} * \text{Volume} \).` First, find the moles of HCl by multiplying its molarity (0.146 moles/L) by its volume (30.4 mL converted to liters). This gives us the moles of HCl present. Since NaOH and HCl react in a 1:1 ratio, the moles of NaOH needed are the same as those of HCl.

Next, using the molarity of NaOH (0.250 moles/L), we can find the volume of NaOH needed by rearranging the equation to
`\( \text{Volume} = \frac{\text{moles}}{\text{Molarity}} \)`. Plug in the moles of NaOH required to neutralize HCl. The result is the volume of NaOH required in liters, which can be converted to milliliters.

In this case, 30.4 mL of 0.146M HCl requires 0.004424 moles of NaOH for neutralization. Using the molarity of NaOH, the volume required is calculated as 27.2 mL. This ensures that there are enough moles of NaOH to react completely with the moles of HCl present, achieving a neutralization reaction.