calculate the molarity of an acetic acid solution if 39.96 mL of the solution is needed to neutralize 136mL of 1.41 M sodium hydroxide. the equation for the reaction is
HC2H3O2(aq) + NaOH(aq) > Na+(aq) + C2H3O2(aq)
+H2O(aq)
mols NaOH = M x L = ?
mols acetic acid = mols NaOH (from the equation which is 1 mol NaOH = 1 mol acetic acid)
M acetic acid = mols acetic acid/L acetic acid.
To find the molarity of the acetic acid solution, we need to use the concept of stoichiometry.
Step 1: Write down the balanced chemical equation for the reaction:
HC₂H₃O₂(aq) + NaOH(aq) → Na⁺(aq) + C₂H₃O₂⁻(aq) + H₂O(l)
Step 2: Identify the stoichiometric ratio between sodium hydroxide (NaOH) and acetic acid (HC₂H₃O₂) from the balanced equation. The stoichiometric ratio is 1:1, which means that one mole of NaOH reacts with one mole of HC₂H₃O₂.
Step 3: Convert the volume of sodium hydroxide solution from mL to L.
136 mL = 136/1000 L = 0.136 L
Step 4: Calculate the number of moles of NaOH using the molarity and volume of NaOH solution:
moles of NaOH = Molarity × Volume in liters
moles of NaOH = 1.41 M × 0.136 L = 0.19176 moles NaOH
Step 5: Since the stoichiometric ratio between NaOH and HC₂H₃O₂ is 1:1, the number of moles of acetic acid is also 0.19176 moles.
Step 6: Convert the volume of acetic acid solution from mL to L.
39.96 mL = 39.96/1000 L = 0.03996 L
Step 7: Calculate the molarity of acetic acid using the number of moles and the volume of acetic acid solution:
Molarity = Moles of solute / Volume of solution in liters
Molarity = 0.19176 moles / 0.03996 L = 4.798 M
Therefore, the molarity of the acetic acid solution is 4.798 M.