1.

20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)_(2) in the sample. (Mol. wt. of Ba(OH_(2))=171)

Answer» <html><body><p></p>Solution :Calculation of volume of HCl used in titration between <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> and HCl. <br/> `V_(NaOH) =10 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>, M_(NaOH) = 0.2 M, M_(HC) = 0.5 M, V_(HCl)`=? <br/> `M_(NaOH) xx V_(NaOH) = M_(HCl) xx V_(HCl), V_(HCl) = (10 xx 0.2)/0.5 = 4 ml` <br/> This is the volume of HCl left unused when excess of HCl is added to `Ba(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(2)` solution. <br/> Total volume of HCl added = 10 ml <br/> Volume of HCl used to react with `Ba(OH)_(2) = 10-4 = 6` ml <br/> `2HCl + Ba(OH)_(2) to BaCl_(2) + 2H_(2)O` <br/> M = No. of moles x `1000/("volume of solution") rArr 0.5 = "moles" xx 1000/6`. <br/> `(0.5 xx 6)/1000` = moles of HCl Moles of HCl used = 0.003 moles. <br/> Observing the molar ratio of HCl and `Ba(OH)_(2)`. <br/> Moles of `Ba(OH)_(2)` <a href="https://interviewquestions.tuteehub.com/tag/reacted-7708680" style="font-weight:bold;" target="_blank" title="Click to know more about REACTED">REACTED</a> `=1/2` x moles of HCl reacted `=1/2 xx 0.003 = 0.0015` moles. <br/> Weight of `Ba(OH)_(2)` reacted = no. of moles x mol. <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>. =` 0.0015 xx 171 = 0.2565` g <br/> Percentage of `Ba(OH)_(2)` in the sample `=(wt. of Ba(OH)_(2) "reacted")/("Total weight of sample") xx 100` = 1.28 %</body></html>


Discussion

No Comment Found

Related InterviewSolutions