Question

A mixture of $NaOH$ and $N{a}_{2}C{O}_3$ solution is titrated with $50\text{mL}$ of $\frac{N}{10}HCl$ using phenolpthalein. At this stage methyl orange was added and addition of acid was continued. The second end point was reached after further addition of $10\text{mL}$ of $\frac{N}{10}HCl$ . The amount of $NaOH\left(\text{in g}\right)$ in the solution is:

• A. $3.2\text{g}$
• B. $0.16\text{g}$
• C. $0.08\text{g}$
• D. $0.4\text{g}$

Answer 1

The correct option is B $0.16\text{g}$

1st titration
$\text{Meq. of}NaOH+\frac{1}{2}\text{Meq. of}N{a}_{2}C{O}_3=\text{Meq. of}HCl=50\times \frac{1}{10}=5$
2nd titration
$\frac{1}{2}\text{Meq. of}N{a}_{2}C{O}_3=\text{Meq. of}HCl=10\times \frac{1}{10}=1$
$\therefore \text{Meq. of}NaOH\text{remains}=5-1=4$
$\text{Wteight of}NaOH=\frac{\text{Meq. of NaOH}\times \text{Equivalent weight}}{1000}=\frac{4\times 40}{1000}\text{g}=0.16\text{g}$