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Question

25 mL of a mixture of NaOH and Na2CO3 when titrated with N/10 HCl using phenolphthalein indicator required 25 mL HCl. The same volume of mixture when titrated with N/10 HCl using methyl orange indicator required 30 mL of HCl. Calculate the amount of Na2CO3 and NaOH in one litre of this mixture.

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Solution

When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e.,
NaOH+HClNaCl+H2O

Na2CO3+HClNaHCO3+NaCl

So, 25 mLN10HClNaOH+1/2Na2CO3 present in 25 mL of mixture.

In another titration, when methyl orange is the indicator, whole of NaOH has been neutralised and carbonate converted into carbonic acid, i.e.,

Na2CO2+2HCl2NaCl+H2CO3

30mLN10HClNaOH+Na2CO3 present in 25 mL of mixture

Hence,
(3025)mLN10HCl12Na2CO3 present in 25 mL of mixture

Hence,
10 mLN10HClNa2CO3 present in 25 mL of mixture

10 mLN10Na2CO3 solution

Amount of Na2CO3=53×1010×1000=0.053 g

This amount of Na2CO3 is present in 25 mL of mixture.

This amount present in one litre of mixture,

=0.05325×1000=2.12 g

(3010)mLN10HClNaOH present in 25 mL of mixture

20 mLN10NaOH

Amount of NaOH in 25 mL of mixture =40×2010×1000=0.08 g

The amount present in one litre of mixture =0.0825×1000=3.20 g.

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