Final answer:
To neutralize 30.4 mL of 0.152 M HCl with 0.250M NaOH, you would need approximately 18.48 mL of NaOH.
Step-by-step explanation:
To calculate the number of milliliters of 0.250M NaOH required to neutralize 30.4 mL of 0.152 M HCl, we can use the concept of stoichiometry and the equation for the neutralization reaction between NaOH and HCl.
First, let's write the balanced chemical equation for the reaction:
HCl + NaOH → NaCl + H₂O
The mole ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaOH to neutralize it.
Now, let's use the equation:
Moles of HCl = volume (L) x concentration (mol/L)
Moles of NaOH = moles of HCl = (30.4 mL / 1000 mL/L) x 0.152 M = 0.0046208 mol
Finally, we can calculate the volume of 0.250M NaOH needed:
Volume (mL) = (moles of NaOH / concentration of NaOH) x 1000 mL
Volume (mL) = (0.0046208 mol / 0.250 M) x 1000 mL = 18.4832 mL