In a neutralization reaction between sodium hydroxide (NaOH) and Hydrochloric acid (HCl); 12.2 mL NaOH is used to neutralize 20 ml 0.32 M HCl solution. Calculate the molarity of the NaOH used?
The Correct Answer and Explanation is:
Correct Answer:
The molarity of NaOH is 0.5246 M (rounded to four decimal places).
Step-by-Step Solution:
This is a neutralization reaction between sodium hydroxide (a strong base) and hydrochloric acid (a strong acid): NaOH+HCl→NaCl+H2O\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}NaOH+HCl→NaCl+H2O
In this balanced chemical equation, the mole ratio is 1:1. That means 1 mole of NaOH reacts with 1 mole of HCl.
Given Data:
- Volume of NaOH = 12.2 mL = 0.0122 L
- Volume of HCl = 20.0 mL = 0.0200 L
- Molarity of HCl = 0.32 M
- Molarity of NaOH = ?
Step 1: Calculate moles of HCl
Moles of HCl=M×V=0.32×0.0200=0.0064 moles\text{Moles of HCl} = M \times V = 0.32 \times 0.0200 = 0.0064 \text{ moles}Moles of HCl=M×V=0.32×0.0200=0.0064 moles
Step 2: Since the reaction is 1:1, moles of NaOH = moles of HCl = 0.0064 moles
Step 3: Use moles and volume of NaOH to find its molarity
Molarity of NaOH=Moles of NaOHVolume in L=0.00640.0122=0.5246 M\text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume in L}} = \frac{0.0064}{0.0122} = 0.5246 \text{ M}Molarity of NaOH=Volume in LMoles of NaOH=0.01220.0064=0.5246 M
Explanation:
Neutralization reactions involve an acid and a base reacting to form a salt and water. In this particular case, both the acid (HCl) and the base (NaOH) are strong, meaning they completely dissociate in water. This simplifies calculations, as we can assume that all the H+^++ ions from the acid react with all the OH−^-− ions from the base.
The most important aspect in solving such problems is recognizing the stoichiometric relationship from the balanced chemical equation. In this 1:1 reaction, the number of moles of acid equals the number of moles of base at the equivalence point. After determining the number of moles of acid used, we directly equate it to the moles of base and use the volume to compute the molarity.
Understanding this concept is fundamental in titration calculations and is widely used in laboratory chemical analysis.
