Question

$40mL$ of $0.05M$ solution of sodium sesquicarbonate $\left(N{a}_{2}C{O}_{3}NaHC{O}_3;2H_{2}O\right)$ is titrated against $0.05MHCl.xmL$ of $HCl$ is used when phenolphthalein is the indicator and $ymL$ of $HCl$ is used when methyl orange is the indicator in two separate titrations. Hence $e(y-x)$ is

• A. $80mL$
• B. $30mL$
• C. $120mL$
• D. None

Answer 1

Answer: (A)

$0.05MN{a}_{2}C{O}_3=1NN{a}_{2}C{O}_3$
$0.05MNaHC{O}_3=0.05NNaHCO$
$40L$ of $0.1N{a}_{2}C{O}_3\equiv 40mL$ of $0.1NHCl$
(For complete reaction) $=80mL$ of $0.05MHCl$
For $50\%$ reaction $=40mL$ of $0.05MHCl$
With phenolphthalein $=x=40mL$
$40mL$ of $0.05NaHC{O}_3=40mL$ of $0.05MHCl$
With methyl orange, $y=80+40=120mL$
$\therefore (y-x)=80mL$