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 0.50 M NaOH, we need to use the balanced chemical equation for the neutralization reaction between HNO3 (nitric acid) and NaOH (sodium hydroxide).
The balanced chemical equation for the reaction is:
HNO3 + NaOH → NaNO3 + H2O
From the balanced equation, we can see that one mole of HNO3 reacts with one mole of NaOH, producing one mole of NaNO3 and one mole of water.
First, we need to determine the number of moles of NaOH in 0.1L of 0.50 M solution:
Number of moles = Volume (L) x Concentration (M)
Number of moles of NaOH = 0.1 L x 0.50 M = 0.05 moles
Since the reaction is 1:1 between HNO3 and NaOH, we conclude that 0.05 moles of HNO3 can be neutralized by 0.05 moles of NaOH.
Therefore, the quantity of HNO3 that can be neutralized by 0.1L of 0.50 M NaOH is 0.05 moles.