Final answer:
To find the volume of 0.250 M NaOH required to neutralize 30.4 mL of 0.146 M HCl, we use stoichiometry of the reaction, which tells us we need an equal number of moles of NaOH as HCl. The calculation gives us 17.8 mL of NaOH required, which doesn't exactly match any of the options provided.
Step-by-step explanation:
To determine how many milliliters of 0.250 M NaOH are required to neutralize 30.4 mL of 0.146 M HCl, we can use the stoichiometry of the neutralization reaction between HCl and NaOH, which is a one-to-one mole ratio:
First, calculate the number of moles of HCl in the 30.4 mL solution:
moles of HCl = Molarity of HCl × Volume of HCl (in liters)
moles of HCl = 0.146 M × 0.0304 L = 0.0044384 mol HCl
Since the mole ratio is 1:1, moles of NaOH needed = moles of HCl.
To find the volume of 0.250 M NaOH needed, we use the molarity (M) equation:
M = moles of solute / volume of solution (in liters)
0.250 M = 0.0044384 mol / volume of NaOH (in liters)
Volume of NaOH = 0.0044384 mol / 0.250 M
Volume of NaOH = 0.0177536 liters
Convert liters to milliliters:
Volume of NaOH = 0.0177536 L × 1000 mL/L = 17.8 mL
Therefore, none of the options provided exactly match the calculation; however, based on the principle of significant figures and the calculation, it appears there may be an error in the provided options or the calculation steps may need to be reviewed.