Question

$40mL$ of $0.05MN{a}_{2}C{O}_3\cdot NaHC{O}_3\cdot 2H_{2}O$ (sesquicarbonate) is titrated against $0.05MHCl.xmL$ of $HCl$ is used when phenolphthalein is the indicator and $ymLHCl$ is used when methyl orange is the indicator in two separate titrations, hence $(y-x)$ is________.

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

Answer 1

The correct option is A $80mL$

Titration of ${Na}_2{CO}_3.{NaHCO}_3.2H_{2}O$ with $HCl$ involves following reactions :

a) ${Na}_2{CO}_3+HCl\rightleftharpoons {NaHCO}_3+NaCl$

b) ${NaHCO}_3+HCl\rightleftharpoons NaCl+H_{2}O+{CO}_2$

In step $a$, $40$ ml of $0.05M$ $HCl$ will react with $40$ ml of $0.05M$ ${Na}_2{CO}_3$ to form ${NaHCO}_3$ using phenolpthalein.

$\therefore$ $x=40$ ml

Now, in a separate titration $40$ ml of $0.05M$ $HCl$ will need to react with $0.05M$ ${Na}_2{CO}_3$ to form ${NaHCO}_3$. Now in the second step $b$ total $80$ ml of $0.05M$ $HCl$ will need to neutralise ${NaHCO}_3$ completely.

$\therefore$ $y=40+40\times 2=120$

$\therefore$ $y-x=120-40=80$ ml

Answer will be $A$.