40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is

40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaH

1.	40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is
Answer» <html><body><p>80 mL<br/>30 mL<br/>120 mL<br/>none of these</p>Solution :In the first <a href="https://interviewquestions.tuteehub.com/tag/titration-710389" style="font-weight:bold;" target="_blank" title="Click to know more about TITRATION">TITRATION</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> = `(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2)` eq `NC_(2)CO_(3)` <br/> `(x xx0.05)/(1000)=(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(0.050xx40xx2)/(1000)impliesx=40ml` <br/> In the second titration <br/> Eq of HCl = Eq `NC_(2)CO_(3)+` Eq `NKHCO_(3)` <br/> `(yxx0.05)/(1000)=(40xx0.05xx2)/(1000)+(40xx0.05xx1)/(1000)` <br/> `y=120impliesy-x=80`</body></html>