What quantity of HNO3 can be neutralized by 0.1L of 0.50 m NaOH
To determine the quantity of HNO3 that can be neutralized by 0.1L of 0.50M NaOH, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HNO3 and NaOH.
The balanced chemical equation for the reaction is:
HNO3 + NaOH -> NaNO3 + H2O
From the balanced equation, we can see that the stoichiometric ratio between HNO3 and NaOH is 1:1. This means that for every 1 mole of HNO3, there is an equal amount of 1 mole of NaOH required for complete neutralization.
First, we need to calculate the number of moles of NaOH in 0.1L of 0.50M NaOH:
Number of moles of NaOH = molarity of NaOH x volume of NaOH
= 0.50M x 0.1L
= 0.05 moles
Since the stoichiometric ratio between HNO3 and NaOH is 1:1, the number of moles of HNO3 required for complete neutralization is also 0.05 moles.
Therefore, 0.1L of 0.50M NaOH can neutralize 0.05 moles of HNO3.