Question

A solution contains a mixture of $N{{a}_{2}}C{{O}_{3}}$ and $NaOH$. Using Ph as an indicator 25mL of mixture required 19.5mL of 0.995N $HCl$ for the end point. With $MeOH$, 25mL of the solution required 25mL of the same $HCl$for the end point. Calculate g/L of each substance in the mixture.

Answer 1

Hint: In the presence of phenolphthalein (Ph) as an indicator, acid will equal half the amount of sodium carbonate and full amount of sodium hydroxide is required. Thus, the answer to this question includes calculation of Meq.

Complete step by step solution:
Let us consider that ‘x’ and ‘y’ are the Meq ( Milliequivalent per litre) of sodium carbonate and sodium hydroxide respectively.
Since, the fact lies that in the presence of phenolphthalein as indicator the acid equals half the amount of $N{{a}_{2}}C{{O}_{3}}$and full of $NaOH$is required then we have,
$\dfrac {1}{2}$ (${M_{eq}}$ $of$ $N{{a}_{2}}C{{O}_{3}}$) + ${M_{eq}}$ $of$ $NaOH$= ${M_{eq}}$ $of$ $HCl$ \[\left( \dfrac{x}{2} \right)+b=19.5\times 0.995\]
\[\left( \dfrac{x}{2} \right)+b=19.5\times 0.995\]
\[\left( \dfrac{x}{2} \right)+b=19.4\] …..(1)
Now , if we use $MeOH$ as an indicator then the acid will equal the amount of sodium carbonate and sodium hydroxide is required. Therefore, the equation becomes,
Meq of $N{{a}_{2}}C{{O}_{3}}$+ Meq of $NaOH$= Meq of $HCl$
\[\Rightarrow x+y=25\times 0.995\]
\[\Rightarrow x+y=24.875\] …..(2)
Subtracting equation (1) from equation (2) , we have
$\left( a+b \right)-\left( \dfrac{a}{2}+b \right)=24.875-19.4$
\[\dfrac{a}{2}=5.475\] or $a=5.475\times 2$
$\Rightarrow a=10.95$
Substituting this above value in the equation number (2) we get,
\[b=24.875-10.95\] $\Rightarrow b=13.925$
Therefore, weight of $N{{a}_{2}}C{{O}_{3}}$per litre of solution = $\dfrac{10.95\times 10H-3}{0.025}\times \dfrac{106}{2}$
$\Rightarrow $ Weight of $N{{a}_{2}}C{{O}_{3}}$=$23.2g{{L}^{-1}}$.
Similarly, weight of $NaOH$per litre of solution will be=$\dfrac{13.925\times {{10}^{-3}}}{0.025}\times \dfrac{40}{1}$
$\Rightarrow $ Weight of $NaOH$=$22.28g{{L}^{-1}}$.

Thus, the amount of each substance that is $N{{a}_{2}}C{{O}_{3}}$ and $NaOH$ are $23.2g{{L}^{-1}}$and $22.28g{{L}^{-1}}$ respectively.

Additional information:
Phenolphthalein is a Redox indicator which is normally used in the acid- base titrations and it is also a pH indicator. It is colourless below 8.5 pH and changes to pink colour or deep red hue above pH=8.5

Note: The quantity Meq ( Milliequivalents per litre) is the measure which is often calculated for electrolytes. This indicates the chemical activity or also the combining power of an element which is relative to the activity of 1mg of ${{H}_{2}}$.