Question

50 mL of 0.5 m oxalic acid is needed to neutralise 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
1. 40 g
2. 80 g
3. 20 g
4. 10 g

Answer 1

Correct Answer - Option 1 : 40 g

Concept:

The reaction takes place as follows,

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

Molarity = \({\rm{\;}}\frac{{{\rm{Number\;of\;moles\;}}}}{{{\rm{Volume\;of\;solution\;}}\left( {{\rm{in\;L}}} \right)}}\)

Calculation:

Now, 50 mL of 0.5 M H₂C₂O₄ is needed to neutralize 25 mL of NaOH

∴ M_eq of H₂C₂O₄ = M_eq of NaOH

50 × 0.5 × 2 = 25 × M_NaOH × 1

M_NaOH = 2M

Now, molarity = \({\rm{\;}}\frac{{{\rm{Number\;of\;moles\;}}}}{{{\rm{Volume\;of\;solution\;}}\left( {{\rm{in\;L}}} \right)}}\)

= \(\frac{{{\rm{weight}}/{\rm{molecular\;mass}}}}{{{\rm{Volume\;of\;solution\;}}\left( {{\rm{in\;L}}} \right)}}\)

2 = \(\frac{{{{\rm{w}}_{{\rm{NaOH}}}}}}{{40}} \times \frac{{1000}}{{50}}\)

\({{\rm{w}}_{{\rm{NaOH}}}} = \frac{{2 \times 40 \times 50{\rm{\;}}}}{{1000}}{\rm{\;}} = 4{\rm{\;g}}.\)

Thus, none of the given options is correct.