Question

$10mL$ of a solution of $NaOH$ is found to be completely neutralised by $8mL$ of a given solution of $HCl$. If we take $20mL$ of the same solution of $NaOH$, the amount $HCl$ solution (the same solution as before) required to neutralise it will be:

• A. $4mL$
• B. $8mL$
• C. $12mL$
• D. $16mL$

Answer 1

The correct option is D $16mL$

Reaction of $HCl$ and $NaOH$ is:

$NaOH+HCl\to NaCl+H_{2}O$

$1mol$ of $NaOH$ is neutralized by $1mol$ of $HCl$. If

$10ml$ of $NaOH$ is neutralize by $HCl=8ml$

$1ml$ of $NaOH$ will neutralize by $HCl=\frac{8}{10}ml$

$20ml$ of $NaOH$ for neutralization will require $HCl=0.8\times 20ml=16ml$

So, the correct option is $D$