To determine the molarity of the sodium hydroxide (base) solution, we can use the concept of stoichiometry and the volume and concentration information provided.
Given:
- Volume of hydrochloric acid (HCl) solution = 25.34 mL
- Concentration of hydrochloric acid (HCl) solution = 6.0 M
- Volume of sodium hydroxide (NaOH) solution = 56.73 mL
- Molarity of sodium hydroxide (NaOH) solution = ?
First, we need to determine the number of moles of HCl used. We can calculate this using the following equation:
moles of HCl = volume of HCl (in liters) × concentration of HCl
Converting the volume of HCl to liters:
25.34 mL = 25.34/1000 = 0.02534 L
Calculating the moles of HCl:
moles of HCl = 0.02534 L × 6.0 M = 0.15204 moles
Since the balanced chemical equation between HCl and NaOH is 1:1, the number of moles of NaOH used will also be 0.15204 moles.
Now, we can determine the molarity of the NaOH solution by dividing the moles of NaOH by the volume in liters:
Molarity of NaOH = moles of NaOH / volume of NaOH (in liters)
Converting the volume of NaOH to liters:
56.73 mL = 56.73/1000 = 0.05673 L
Calculating the molarity of NaOH:
Molarity of NaOH = 0.15204 moles / 0.05673 L ≈ 2.68 M
Therefore, the molarity of the sodium hydroxide (base) solution is approximately 2.68 M.