0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be

0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution

1.	0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a><br/>5<br/>8<br/><a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a></p>Solution :`0.365 g <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> =(0.365)/(36.5) "mole" = 0.01 "mole"` <br/> `100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100` <br/> `=0.02 ` mole <br/> NaOH left unneutralized = 0.01 mole <br/> Volume of solution = 100 ml <br/> `:.` Molarity of NaOH in the solution<br/> `=(0.01)/(100)xx1000=0.1M = 10^(-1)M` <br/> `:. [<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13`</body></html>