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  2. 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution.
# Engineering

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 :

Answers (1)
S Satyajeet Kumar

aA (g) + bB (g) → cC (g) + dD (g)

Here, ‘a’ moles of A(g) reacts with ‘b’ moles of B(g) to give ‘c’ mole of C(g) and ‘d’ moles of D(g)

As we have learned from mole concept

H_{2}C_{2}O_{4}+2NaOH\rightarrow Na_{2}C_{2}O_{4}+2H_{2}O

meq\, \, \, of H_{2}C_{2}O_{4}=meq\, \, NaOH

50\times 0.5\times 2=25MnaOH\times 1

1000ml\, \, solution= 2\times 240gram NaOH

\therefore 50ml \, \, sol=4g \, \, NaOH

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