Question

A solution containing $N{a}_{2}C{O}_3$ and $NaOH$ requires $300ml$ of $0.1NHCl$ using phenolphthalein as indicator. Methyl orange is then added to the above solution when $25ml$ of $0.2NHCl$ is required. The amount of $NaOH$ present in the solution is:

• A. $1.0$ g
• B. $2.0$ g
• C. $1.5$ g
• D. $0.2$ g

Answer 1

Answer: (A)

$1.0$ g

The reaction are as follows:

$NaOH+HCl\to NaCl+H_{2}O$
$N{a}_{2}C{O}_3+HCl\to 2NaCl+C{O}_2+H_{2}O$

Phenolphthalein changes its color when the solution becomes neutral i.e. when all the base is consumed and half of the sodium carbonate is converted to sodium hydrogen carbonate.
So, we can say, $300$ ml $HCl$ of $0.1N$ $HCl$ neutralizes the entire amount of $NaOH$ and half of $N{a}_{2}C{O}_3$.
As given, $50ml$ of $0.1N$ (which is equal to $25$ ml of $0.2N$) of $HCl$ is utilized to neutralize the remaining half of $N{a}_{2}C{O}_3$.
So, $250$ ml ($300-50$ ml) of $0.1N$ of $HCl$ is required to neutralize $NaOH$ completely.

As number of milliequivalents of $HCl=$ number of milliequivalents of $NaOH$
$\Rightarrow 250\times 0.1$ meq of $HCl=250\times 0.1$ meq of $NaOH$ which is equal to $25$ milimoles of $NaOH,$ which in turn, is equal to $1$ g of $NaOH$.