Question

Volume of 0.1 M NaOH required to completely neutralise $40c{m}^3$ of 0.1 M oxalic acid is

• A. 40 cc
• B. 80 cc
• C. 120 cc
• D. 20 cc

Answer 1

The correct option is B 80 cc

The normality of 0.1 M oxalic acid,
$N_A=0.1\times 2=0.2N$
The volume of 0.1 M oxalic acid,
$V_A=40cc$
The normality of 0.1 M NaOH,
$N_B=0.1\times 1=0.1N$
Volume of 0.1 M NaOH,$V_B$
For neutralization,
$N_A{V}_A=N_B{V}_B$
$0.2\times 40=0.1\times V_B$
Solving, $V_B=80cc$.