Question

$40$ g $NaOH$, $106$ g $N{a}_{2}C{O}_3$ and $84$ g $NaHC{O}_3$ are dissolved in water and the solution is made up to $1$ litre. $20$ ml of this stock solution is titrated with $1$ M $HCl$. Which of the following statements are correct?

• A. The titre reading of $HCl$ will be $40$ ml if phenolphthalein is used as an indicator from the very beginning.
• B. The titre reading of $HCl$ will be $60$ ml if phenolphthalein is used as an indicator from the very beginning.
• C. The titre reading of $HCl$ will be $40$ ml if methyl orange is used as an indicator after the $1st$ endpoint.
• D. The titre reading of $HCl$ will be $80$ ml if methyl orange is used as an indicator from the very beginning.

Answer 1

Answer: (B), (C)

The correct options are
B The titre reading of $HCl$ will be $60$ ml if phenolphthalein is used as an indicator from the very beginning.
C The titre reading of $HCl$ will be $40$ ml if methyl orange is used as an indicator after the $1st$ endpoint.

Milli-equivalents of $N{a}_{2}C{O}_3$, $NaHC{O}_3$ and $NaOH$ in $20$ ml solution are $40,20$ and $20$ respectively.
When phenolphthalein is used from the beginning then,
meq. of HCl $=$ meq. of NaOH + meq. of $N{a}_{2}C{O}_3$
Therefore, $1\times 1\times V=(1\times 20\times 1)+(1\times 20\times 2)$.
Hence, $V=60$ ml.

When methyl orange is used from beginning then,
meq. of HCl $=$ (meq. of NaOH + meq. of $NaHC{O}_3$ + $2\times$ meq. of $N{a}_{2}C{O}_3$
$1\times 1\times V=(1\times 20\times 1)+(1\times 20\times 1)+(2\times 1\times 20\times 2)$
Hence, $V=120$ ml.

When methyl orange is used after the first end point then,
meq. of HCl $=2\times$ meq. of $NaHC{O}_3$
$1\times 1\times V=(2\times 1\times 20\times 1)$
Hence, $V=40$ ml.