Question

$40$ ml of $0.05M$ solution of sesqui-carbonate $(N{a}_{2}C{O}_3.NaHC{O}_3.2H_{2}O)$ is titrated against $0.05M$ $HCl$ . $x$ mL of $HCl$ is used when phenolphthalein is the indicator and $y$ mL of $HCl$ is used when methyl orange is the indicator in two separate titrations. Hence, $(y-x)$ is :

• A. $80$ mL
• B. $30$ mL
• C. $120$ mL
• D. none of the above

Answer 1

The correct option is B $30$ mL

The ratio of sodium carbonate and sodium bicarbonate is $1:1$.
Phenolphthalein indicator shows end point corresponding to $50\%$ neutralization of sodium carbonate to sodium bicarbonate.
Methyl orange indicator shows end point corresponding to $100\%$ percent neutralization of sodium carbonate. It also shows neutralization of sodium bicarbonate.
$40$ mL of $0.05MN{a}_{2}C{O}_3.NaHC{O}_3.2H_{2}O$ corresponds to $2$ millimoles.
The number of millimoles of sodium carbonate and sodium bicarbonate are $1$ millimole each.
For $50\%$ neutralization of $1$ millimoles of sodium carbonate, the millimoles of $HCl$ required is $0.05x$ millimoles.
This corresponds to $0.5$ millimoles
For methyl orange indicator, the millimoles of $HCl$ used is $0.05y$ millimoles.
This corresponds to $2$ millimoles
Hence, $x=10$ and $y=40$
$y-x=40-10=30$ mL