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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.

Answer» <html><body><p></p>Solution :When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e., <br/> `NaOH+<a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> to NaCl+H_(2)O` <br/> `Na_(2)CO_(3)+HCl to NaHCO_(3)+NaCl` <br/> So, `<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mL (N)/(10)HCl-=NaOH+1//2Na_(2)CO_(3)` present in 25 mL of mixture <br/> In <a href="https://interviewquestions.tuteehub.com/tag/another-876628" style="font-weight:bold;" target="_blank" title="Click to know more about ANOTHER">ANOTHER</a> titration when methyl orange is the indicator, whole of NaOH has been neutralised and carbonate converted into carbonic acid, i.e., <br/> `Na_(2)CO_(3)+2HCl to 2NaCl +H_(2)CO_(3)` <br/> `30 mL(N)/(10) HCl -=NaOH+Na_(2)CO_(3)` present in 25 mL of mixture <br/> Hence, <br/> `(30-25)mL (N)/(10)HCl-=(1)/(2)Na_(2)CO_(3)` present in 25 mL of mixture <br/> Hence, <br/> `10 mL (N)/(10)HCl-=Na_(2)CO_(3)` present in 25 mL of mixture<br/> `-=10 mL(N)/(10)Na_(2)CO_(3)` solution <br/> Amount of `Na_(2)CO_(3)=(53xx10)/(10xx1000)=0.053 g` <br/> This amount of `Na_(2)CO_(3)` is present in 25 mL of mixture. <br/> The amount present in one litre of mixture <br/> `=(0.053)/(25)xx1000=2.12g` <br/> `(30-10)mL (N)/(10)HCl-=NaOH` present in 25 mL of mixture <br/> `-=20 mL(N)/(10)NaOH` <br/> Amount of NaOH in 25 mL of mixture`=(40xx20)/(10xx1000)=0.08 g` <br/> The amount present in one litre of mixture`=(0.08)/(25)xx1000=3.20 g`</body></html>


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