Final answer:
The molarity of the NaOH solution is determined to be 1.20 M by using the stoichiometry of the titration reaction and the volumes of HCl and NaOH involved, where the reaction has a 1:1 molar ratio between HCl and NaOH.
Step-by-step explanation:
To determine the molarity of NaOH solution, we will use the concept of titration and the stoichiometry of the reaction. Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react in a 1:1 molar ratio, as shown in the balanced equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
The molarity of NaOH can be calculated using the volume and molarity of HCl that was required to neutralize it. As the balanced equation shows a 1:1 stoichiometry, the moles of NaOH will be equal to the moles of HCl used.
To find the moles of HCl:
# moles HCl = Volume HCl (L) × Molarity HCl (M)
# moles HCl = 0.025 L × M
Since the volume of HCl is given as 25 mL, we need to convert this to liters (L) by dividing by 1000 mL/L (25 mL ÷ 1000 mL/L = 0.025 L). Now we can calculate the moles of NaOH which will be the same as the moles of HCl since they react 1:1:
# moles NaOH = # moles HCl = 0.025 L × M
To find the molarity of the NaOH solution, use the formula:
Molarity NaOH = moles NaOH / Volume NaOH (L)
We have the volume of NaOH solution as 30 mL, which is 0.030 L. Using the equation:
Molarity NaOH = (0.025 L × M) / 0.030 L
By solving for M, we find that the molarity of NaOH is 1.20 M, which corresponds to answer choice (B).