How many grams of sodium hydroxide is needed to react with 100 ml of a 1.4 M nitric acid for complete neutralization to happen.
NaOH + HNO3 ==> NaNO3 + H2O
mols HNO3 = M x L = ?
mols NaOH = mols HNO3 (look at the coefficients in the balanced equation.)
mols NaOH = grams/molar mass
39.997 g/mol
To determine the number of grams of sodium hydroxide needed to react with 100 ml of a 1.4 M nitric acid, we need to use the concept of stoichiometry.
First, we need to balance the chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and nitric acid (HNO3):
HNO3 + NaOH → NaNO3 + H2O
From the balanced equation, we can see that the ratio of sodium hydroxide to nitric acid is 1:1. This means that for every 1 mole of nitric acid, we need 1 mole of sodium hydroxide for complete neutralization.
Now, let's calculate the number of moles of nitric acid using its molarity and volume:
Molarity of nitric acid (HNO3) = 1.4 M
Volume of nitric acid = 100 ml = 0.1 L (since 1 ml = 0.001 L)
Number of moles of nitric acid = Molarity × Volume
= 1.4 M × 0.1 L
= 0.14 moles
Since we need an equal amount of sodium hydroxide for neutralization, we require 0.14 moles of sodium hydroxide.
Finally, to calculate the grams of sodium hydroxide needed, we need to use its molar mass. The molar mass of sodium hydroxide (NaOH) is approximately 40 g/mol.
Mass of sodium hydroxide = Number of moles × Molar mass
= 0.14 moles × 40 g/mol
= 5.6 grams
Therefore, you would need approximately 5.6 grams of sodium hydroxide to completely neutralize 100 ml of a 1.4 M nitric acid.