Question

$50\text{mL}$ solution having $NaOH$ and $N{a}_{2}C{O}_3$ require, $0.5M10\text{mL}HCl$ for complete titration with phenolphthalein indicator and another $50\text{mL}$ solution require $0.5M,12\text{mL}HCl$ solution in methyl orange indicator then what is the ratio of number of mole of $NaOH$ and $N{a}_{2}C{O}_3$?

• A. $0.25$
• B. $4$
• C. $2$
• D. $3$

Answer 1

The correct option is B $4$

Let, the mass of $NaOH$ be $x$
The mass of $N{a}_{2}C{O}_3$ be $y$
In case of phenolphthalein, the reactions are:
$NaOH+HCl\to NaCl+H_{2}ON{a}_{2}C{O}_2+HCl\to NaHC{O}_3+HCl$
So here,
Equivalents of $NaOH$ + equivalents of $N{a}_{2}C{O}_3$ =equivalents of $HCl$
So,
$\frac{x}{40}+\frac{y}{106}=\mathrm{0.005...}\left(i\right)$
In case of methyl orange indicator, the reactions are:
$NaOH+HCl\to NaCl+H_{2}ON{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}C{O}_3$
So, equating the equivalents,
$\frac{x}{40}+\frac{\frac{y}{106}}{2}=\mathrm{0.006...}\left(ii\right)$
$\Rightarrow 53x+20y=\mathrm{10.6...}\left(iii\right)$
$\Rightarrow 53x+40y=\mathrm{12.72...}\left(iv\right)$
$\left(iv\right)-\left(iii\right)$, we get
$20y=2.12\Rightarrow y=0.106$
$x=\frac{10.6-2.12}{53}=0.16$
$\therefore$ Moles of $N{a}_{2}C{O}_3=\frac{0.106}{106}=0.001$
Moles of $NaOH=\frac{0.16}{40}=0.004$
So,
$\frac{\text{Moles of NaOH}}{\text{Moles of}N{a}_{2}C{O}_3}=\frac{4}{1}=4$