Question

$20$ mL of $0.1M$ solution of compound $N{a}_{2}C{O}_3.NaHC{O}_3.2H_{2}O$, is titrated against $0.05MHCl.x$ mL of $HCl$ is used when phenolphthalein is used as an indicator and $y$ mL of $HCl$ is used when methyl orange is the indicator in two separate titrations. Hence, $(y-x)$ is:

• A. $40$ mL
• B. $80$ mL
• C. $120$ mL
• D. none of the above

Answer 1

The correct option is B $80$ mL

When phenolphthalein is used as an indicator,
$0.05x=20\times 0.1\times 1;x=40$ mL (Here, only one part reacts)
When methyl orange is used as an indicator;
$0.05y=20\times 0.1\times 3$ (Here, all three parts of the base reacts)
or $y=120$ mL
$\therefore y-x=80$ mL