What is the molarity of a nitric acid solution if 43.33 mL of 0.10000 KOH solution is needed to neutralize 20.00 mL of the acid?
mols KOH = M x L = ?
mols HNO3 = mols KOH (look at the coefficients.)
M HNO3 = mols HNO3/L HNO3.
0.213
To find the molarity of the nitric acid solution, we can use the concept of stoichiometry.
First, let's determine the number of moles of KOH solution used to neutralize the acid.
Given:
Volume of KOH solution = 43.33 mL
Molarity of KOH solution = 0.10000 M
Using the formula:
moles = molarity x volume (in liters)
moles of KOH = 0.10000 M x (43.33 mL / 1000 mL/L)
= 0.10000 M x 0.04333 L
= 0.004333 moles
According to the balanced chemical equation between KOH and HNO3:
KOH + HNO3 ⟶ KNO3 + H2O
The stoichiometric ratio is 1:1, meaning 1 mole of KOH reacts with 1 mole of HNO3.
Since the number of moles of KOH is equal to the number of moles of HNO3, the molarity of the nitric acid solution can be calculated.
Volume of nitric acid solution = 20.00 mL
Using the formula:
molarity = moles / volume (in liters)
molarity of nitric acid solution = (0.004333 moles) / (20.00 mL / 1000 mL/L)
= (0.004333 moles) / (0.02000 L)
= 0.2167 M
Therefore, the molarity of the nitric acid solution is approximately 0.2167 M.