To determine the weight of sodium hydroxide (NaOH) required to neutralize 100 ml of 0.1N hydrochloric acid (HCl), we can follow the steps outlined below:
Understanding the Reaction
- Sodium hydroxide (NaOH) is a strong base, and hydrochloric acid (HCl) is a strong acid.
- The neutralization reaction can be represented as:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Calculating Moles of HCl
- Normality (N) is defined as the equivalent concentration. For HCl, which has one ionizable hydrogen, 1N is equivalent to 1M.
- Volume of HCl = 100 ml = 0.1 L
- Therefore, moles of HCl:
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume} = 0.1 \times 0.1 = 0.01 \text{ moles} \]
Moles of NaOH Required
- The stoichiometry of the reaction shows that 1 mole of NaOH neutralizes 1 mole of HCl.
- Thus, moles of NaOH required = 0.01 moles.
Calculating Weight of NaOH
- The molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
- Weight of NaOH required:
\[ \text{Weight} = \text{Moles} \times \text{Molar Mass} = 0.01 \times 40 = 0.4 \text{ grams} \]
Conclusion
- To neutralize 100 ml of 0.1N HCl, you will need 0.4 grams of sodium hydroxide (NaOH).