In the following acid—base neutralization reaction
HNO3(aq)+ KOH(aq) ----KNO3(aq) + H2O (I)
What is the molarity (M) of an HNO3 solution if 50.0 ml is needed to react with 25.0 ml of 0.150M KOH solution?
mols KOH = M x L = ?
Look at the equation; it is 1:1, therefore, mols HNO3 = mols KOH.
M HNO3 = mols HNO3/L HNO3.
To find the molarity (M) of an HNO3 solution, you need to use the concept of stoichiometry and the balanced equation of the reaction.
Given:
Volume of HNO3 solution = 50.0 mL
Volume of KOH solution = 25.0 mL
Molarity of KOH solution = 0.150 M
From the balanced equation:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
The balanced equation tells us that the ratio of HNO3 to KOH is 1:1. So, for every 1 mole of HNO3, you need 1 mole of KOH to react completely.
Step 1: Convert the given volumes to liters.
Volume of HNO3 solution = 50.0 mL = 50.0 mL × (1 L/1000 mL) = 0.0500 L
Volume of KOH solution = 25.0 mL = 25.0 mL × (1 L/1000 mL) = 0.0250 L
Step 2: Calculate the number of moles of KOH using its molarity and volume.
Moles of KOH = Molarity × Volume (in liters)
Moles of KOH = 0.150 M × 0.0250 L = 0.00375 mol
Step 3: Since the stoichiometry of the balanced equation is 1:1, the number of moles of HNO3 will also be 0.00375 mol.
Step 4: Calculate the molarity of the HNO3 solution.
Molarity = Moles of solute / Volume of HNO3 solution (in liters)
Molarity = 0.00375 mol / 0.0500 L = 0.075 M
Therefore, the molarity of the HNO3 solution is 0.075 M.