Question

What amount of water is added in $40mL$ of $NaOH(0.1N)$ which is neutralized by $50mL$ of $HCl(0.2N)?$

• A. 80 mL
• B. 60 mL
• C. 40 mL
• D. 90 mL

Answer 1

Answer: (B)

(i) $N_1{V}_1=N_2{V}_2$
(ii) Amount of water to be added$=$ total volume $-$ volume of $NaOH$
Given normality of $NaOH=N_1=0.1N$
Volume of $NaOH=V_1=?$
Normality of $HCl{N}_2-0.2N$
Volume of $HCl{V}_2=50mL$
$N_1{V}_1=N_2{V}_2$
$0.1\times V_1=0.2\times 50$
$V_1=\frac{0.2\times 50}{0.1}=100mL$
$V$ of $NaOH=40mL$
Amount $H_{2}O$ to be added $=100-40=60mL$