1. Tardigrade
  2. Question
  3. Chemistry
  4. 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $50\, mL$ of $0.5\, M$ oxalic acid is needed to neutralize $25\, mL$ of sodium hydroxide solution. The amount of $NaOH$ in $50\, mL$ of the given sodium hydroxide solution is :

JEE MainJEE Main 2019Some Basic Concepts of Chemistry

Solution:

$ H _2 C _2 O _4+2 NaOH \longrightarrow Na _2 C _2 O _4+2 H _2 O$
$ m _{ eq } \text { of } H _2 C _2 O _4= m _{ eq } NaOH $
$50 \times 0.5 \times 2=25 \times M_{ NaOH } \times 1$
$\therefore M _{ NaOH }=2 M$
Now $1000 ml$ solution $=2 \times 40$ gram $NaOH$
$\therefore 50 ml$ solution $=4$ gram $NaOH$