Volume of 0.1 textM textH2 textS textO4 required to neutralize 30 textm textL of 0.2 textN textN texta textO textH is

Question

Volume of 0.1 textM textH2 textS textO4 required to neutralize 30 textm textL of 0.2 textN textN texta textO textH is (A) 30 mL (B) 15 mL (C) 40 mL (D) 60 mL

Tardigrade Admin · Accepted Answer

0.1 textM texto textf textH2 textS textO4 ⇒ 0.2 textN texto textf textH2 textS textO4 ∴ textN1 textV1= textN2 textV2 [ textN1 = 0.2 textN textH2 textSO4] 0.2 × textV1=30 × 0.2 ∴ textV1=30 textm textL

Q. Volume of $0.1 M$ $H_{2} S O_{4}$ required to neutralize $30 m L$ of $0.2 N$ $N a O H$ is

$30 m L$

$15 m L$

$40 m L$

$60 m L$

$0.1 M o f H_{2} S O_{4} \Rightarrow 0.2 N o f H_{2} S O_{4}$

$∴$ $N_{1} V_{1} = N_{2} V_{2}$ $[N_{1} = 0.2 N H_{2} SO_{4}]$

$0.2 \times V_{1} = 30 \times 0.2$

$∴$ $V_{1} = 30 m L$

Q. Volume of 0.1M H2​SO4​ required to neutralize 30mL of 0.2N NaOH is

30mL

15mL

40mL

60mL