Question

A solution contains a mixture of $KOH$ and $N{a}_{2}C{O}_3$. This solution requires $15$ ml of $\frac{N}{20}HCl$ when titrated with phenolphthalein as an indicator. However, the same amount of the solution when titrated with methyl orange as an indicator requires $25$ ml of the same acid. If the summation of weights of $KOH$ and $N{a}_{2}C{O}_3$ in the solution is $x$ g,

then find the value of $\frac{13.4}{100x}$.

(Given that the molecular mass of $KOH$ is $56.1$ g/mol and of $N{a}_{2}C{O}_3$ is $106$ g/mol.)

Answer 1

Given that a solution contains a mixture of $KOH$ and $N{a}_{2}C{O}_3$.

During titration with phenolphthalein indicator, complete neutralisation of $KOH$ and half neutralisation of sodium carbonate occurs.
During titration with methyl orange indicator, complete neutralisation of $KOH$ and complete neutralisation of sodium carbonate occurs.
Hence, the difference in the volume of $HCl$ required for titration with methyl orange and the titration with phenolphthalein corresponds to half neutralisation of sodium carbonate.
It is equal to $25-15=10$ mL of $\frac{N}{20}$ $HCl$ solution.
The number of moles of $HCl$ and hence, the number of moles of sodium carbonate is $\frac{10}{1000\times 20}=0.0005$ moles.
The molar mass of sodium carbonate is $106$ g/mol.
Hence, the mass of sodium carbonate present in the sample is $106\times 0.0005=0.053$ g
For titration with phenolphthalein indicator, $15$ ml of $HCl$ are required out of which $10$ ml corresponds to half neutralisation of sodium carbonate and $5$ ml corresponds to neutralisation of $KOH$.
Hence, the number of moles of $KOH$ present are equal to one half the number of moles of sodium carbonate.

It is equal to $\frac{0.0005}{2}=0.00025$ moles
The molar mass of $KOH$ is $56.1$ g/mol.

The mass of $KOH$ is $56.1\times 0.00025=0.014$ g.
Total mass of the mixture is $0.053+0.014=0.067$ g

It is equal to $x$ g.
Hence, the value of $\frac{13.4}{100x}$ is $\frac{13.4}{100\times 0.067}=2$