To determine the weight of sodium hydroxide (NaOH) required to neutralize 100 ml of 0.1N hydrochloric acid (HCl), we can follow these steps:
Understanding Normality
- Normality (N) is a measure of concentration equivalent to molarity but considers the reactive capacity of the solute.
- HCl is a strong acid that dissociates completely in solution, and it has one acidic hydrogen, thus 1N HCl means that 1 L of solution contains 1 equivalent of HCl.
Calculating Moles of HCl
- Volume of HCl = 100 ml = 0.1 L
- Normality of HCl = 0.1N
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Moles of HCl} = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]
Neutralization Reaction
- The balanced chemical equation for the reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- From the equation, 1 equivalent of HCl reacts with 1 equivalent of NaOH.
Calculating Weight of NaOH
- Since 0.01 equivalents of HCl will require 0.01 equivalents of NaOH,
- The molar mass of NaOH = 40 g/mol.
\[ \text{Weight of NaOH} = \text{equivalents} \times \text{molar mass} \]
\[ \text{Weight of NaOH} = 0.01 \, \text{equiv.} \times 40 \, \text{g/mol} = 0.4 \, \text{g} \]
Conclusion
- To neutralize 100 ml of 0.1N HCl, you will need 0.4 grams of sodium hydroxide (NaOH).