how much of naoh is required to neutralize 1500 ml of 0.1 N hcl
To determine the amount of NaOH required to neutralize 1500 mL of 0.1 N HCl, you can use the concept of stoichiometry. The balanced chemical equation for the reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid) is:
NaOH + HCl → NaCl + H2O
Based on the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl.
First, let's determine the number of moles of HCl in 1500 mL of 0.1 N HCl:
moles of HCl = volume (in liters) × concentration (in moles per liter)
= 1.5 L × 0.1 mol/L
= 0.15 moles
From the balanced equation, we know that the stoichiometric ratio of NaOH to HCl is 1:1. Therefore, the number of moles of NaOH required to neutralize the given amount of HCl is also 0.15 moles.
Finally, we convert the moles of NaOH to the required volume in milliliters (mL) using the concentration of NaOH.
volume (in liters) = moles / concentration (in moles per liter)
= 0.15 mol / 1 mol/L
= 0.15 L
Now, convert the volume from liters to milliliters:
volume (in mL) = volume (in liters) × 1000
= 0.15 L × 1000
= 150 mL
Therefore, 150 mL of 0.1 N NaOH is required to neutralize 1500 mL of 0.1 N HCl.