50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is

50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The tit

1.	50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is
Answer» <html><body><p>`12 cm^(3)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> cm^(3)`<br/>`25 cm^(3)`<br/>`10.5cm^(3)`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> cm^(3) ` of 0.2 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</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> `=50xx0.2` millieq <br/> =10 millieq <br/> `50 cm^(3)` of 0.1 N NaOH = 5 millieq <br/> Remaining HCl = 10 - 5 millieq = 5 millieq <br/> It is neutralized by `V cm^(3)` of 0.5 N KOH i.e. `0.5xxV` millieq = 5 millieq <br/> or `V=10 cm^(3)`.</body></html>