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25 mL of a mixture of NaOH and `Na_(2)CO_(3)` when itrated 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 `Na_(2)CO_(3)` and NaOH in one litre of this mixture. |
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Answer» When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e., `NaOH+HCl to NaCl+H_(2)O` `Na_(2)CO_(3)+HCl to NaHCO_(3)+NaCl` So, `25 mL (N)/(10)HCl-=NaOH+1//2Na_(2)CO_(3)` 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., `Na_(2)CO_(3)+2HCl to 2NaCl +H_(2)CO_(3)` `30 mL(N)/(10) HCl -=NaOH+Na_(2)CO_(3)` present in 25 mL of mixture Hence, `(30-25)mL (N)/(10)HCl-=(1)/(2)Na_(2)CO_(3)` present in 25 mL of mixture Hence, `10 mL (N)/(10)HCl-=Na_(2)CO_(3)` present in 25 mL of mixture `-=10 mL(N)/(10)Na_(2)CO_(3)` solution Amount of `Na_(2)CO_(3)=(53xx10)/(10xx1000)=0.053 g` This amount of `Na_(2)CO_(3)` is present in 25 mL of mixture. The amount present in one litre of mixture `=(0.053)/(25)xx1000=2.12g` `(30-10)mL (N)/(10)HCl-=NaOH` present in 25 mL of mixture `-=20 mL(N)/(10)NaOH` Amount of NaOH in 25 mL of mixture`=(40xx20)/(10xx1000)=0.08 g` The amount present in one litre of mixture`=(0.08)/(25)xx1000=3.20 g` |
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