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Question: A solution contains a mixture of \(N{{a}_{2}}C{{O}_{3}}\) and \(NaOH\). Using Ph as an indicator 25m...

A solution contains a mixture of Na2CO3N{{a}_{2}}C{{O}_{3}} and NaOHNaOH. Using Ph as an indicator 25mL of mixture required 19.5mL of 0.995N HClHCl for the end point. With MeOHMeOH, 25mL of the solution required 25mL of the same HClHClfor the end point. Calculate g/L of each substance in the mixture.

Explanation

Solution

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 Na2CO3N{{a}_{2}}C{{O}_{3}}and full of NaOHNaOHis required then we have,
12\dfrac {1}{2} (Meq{M_{eq}} ofof Na2CO3N{{a}_{2}}C{{O}_{3}}) + Meq{M_{eq}} ofof NaOHNaOH= Meq{M_{eq}} ofof HClHCl (x2)+b=19.5×0.995\left( \dfrac{x}{2} \right)+b=19.5\times 0.995
(x2)+b=19.5×0.995\left( \dfrac{x}{2} \right)+b=19.5\times 0.995
(x2)+b=19.4\left( \dfrac{x}{2} \right)+b=19.4 …..(1)
Now , if we use MeOHMeOH as an indicator then the acid will equal the amount of sodium carbonate and sodium hydroxide is required. Therefore, the equation becomes,
Meq of Na2CO3N{{a}_{2}}C{{O}_{3}}+ Meq of NaOHNaOH= Meq of HClHCl
x+y=25×0.995\Rightarrow x+y=25\times 0.995
x+y=24.875\Rightarrow x+y=24.875 …..(2)
Subtracting equation (1) from equation (2) , we have
(a+b)(a2+b)=24.87519.4\left( a+b \right)-\left( \dfrac{a}{2}+b \right)=24.875-19.4
a2=5.475\dfrac{a}{2}=5.475 or a=5.475×2a=5.475\times 2
a=10.95\Rightarrow a=10.95
Substituting this above value in the equation number (2) we get,
b=24.87510.95b=24.875-10.95 b=13.925\Rightarrow b=13.925
Therefore, weight of Na2CO3N{{a}_{2}}C{{O}_{3}}per litre of solution = 10.95×10H30.025×1062\dfrac{10.95\times 10H-3}{0.025}\times \dfrac{106}{2}
\Rightarrow Weight of Na2CO3N{{a}_{2}}C{{O}_{3}}=23.2gL123.2g{{L}^{-1}}.
Similarly, weight of NaOHNaOHper litre of solution will be=13.925×1030.025×401\dfrac{13.925\times {{10}^{-3}}}{0.025}\times \dfrac{40}{1}
\Rightarrow Weight of NaOHNaOH=22.28gL122.28g{{L}^{-1}}.

Thus, the amount of each substance that is Na2CO3N{{a}_{2}}C{{O}_{3}} and NaOHNaOH are 23.2gL123.2g{{L}^{-1}}and 22.28gL122.28g{{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 H2{{H}_{2}}.