Understanding Neutralization
Neutralization is a chemical reaction between an acid and a base, resulting in the formation of water and a salt. In this case, we need to neutralize hydrochloric acid (HCl) with sodium hydroxide (NaOH).
Given Data
- Volume of HCl solution = 100 ml = 0.1 L
- Normality of HCl = 0.1 N
Calculating Moles of HCl
- Moles of HCl = Normality × Volume (in L)
- Moles of HCl = 0.1 N × 0.1 L = 0.01 equivalents
Since HCl is a strong acid and dissociates completely, it reacts in a 1:1 ratio with NaOH. Thus, 0.01 equivalents of NaOH are required for neutralization.
Calculating Weight of NaOH
1. Equivalent weight of NaOH
- Equivalent weight = Molecular weight / n
- Molecular weight of NaOH = 40 g/mol
- Since NaOH donates one hydroxide ion (n = 1), equivalent weight = 40 g/equiv.
2. Weight of NaOH required
- Weight = equivalents × equivalent weight
- Weight = 0.01 equiv × 40 g/equiv = 0.4 g
Conclusion
To neutralize 100 ml of 0.1N HCl, 0.4 grams of sodium hydroxide is required. This calculation demonstrates the stoichiometric relationship between the acid and base involved in the neutralization reaction.