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

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 : (A) 40 g (B) 20 g (C) 80 g (D) 4 g

Tardigrade Admin · Accepted Answer

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 × 0.5 × 2=25 × M NaOH × 1 ∴ M NaOH =2 M Now 1000 ml solution =2 × 40 gram NaOH ∴ 50 ml solution =4 gram NaOH

Q. $50 m L$ of $0.5 M$ oxalic acid is needed to neutralize $25 m L$ of sodium hydroxide solution. The amount of $N a O H$ in $50 m L$ of the given sodium hydroxide solution is :

40 g

20 g

80 g

4 g

$H_{2} C_{2} O_{4} + 2 N a O H ⟶ N a_{2} C_{2} O_{4} + 2 H_{2} O$
$m_{e q} of H_{2} C_{2} O_{4} = m_{e q} N a O H$
$50 \times 0.5 \times 2 = 25 \times M_{N a O H} \times 1$
$∴ M_{N a O H} = 2 M$
Now $1000 m l$ solution $= 2 \times 40$ gram $N a O H$
$∴ 50 m l$ solution $= 4$ gram $N a O H$

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

40 g

20 g

80 g

4 g

H2​C2​O4​+2NaOH⟶Na2​C2​O4​+2H2​O meq​ of H2​C2​O4​=meq​NaOH 50×0.5×2=25×MNaOH​×1 ∴MNaOH​=2M Now 1000ml solution =2×40 gram NaOH ∴50ml solution =4 gram NaOH

Q. $50 m L$ of $0.5 M$ oxalic acid is needed to neutralize $25 m L$ of sodium hydroxide solution. The amount of $N a O H$ in $50 m L$ of the given sodium hydroxide solution is :

$H_{2} C_{2} O_{4} + 2 N a O H ⟶ N a_{2} C_{2} O_{4} + 2 H_{2} O$
$m_{e q} of H_{2} C_{2} O_{4} = m_{e q} N a O H$
$50 \times 0.5 \times 2 = 25 \times M_{N a O H} \times 1$
$∴ M_{N a O H} = 2 M$
Now $1000 m l$ solution $= 2 \times 40$ gram $N a O H$
$∴ 50 m l$ solution $= 4$ gram $N a O H$