2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@) C. The concentration of H^(+) at equivalence point is (K_(w) = 1 xx 10^(-14)"at"25^(@)C)

2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)

1.	2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@) C. The concentration of H^(+) at equivalence point is (K_(w) = 1 xx 10^(-14)"at"25^(@)C)
Answer» <html><body><p>`3.7xx10^(-13)M`<br/>`3.2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>) M`<br/>`3.2xx10^(-2)M`<br/>`2.7xx10^(-2)M`</p>Solution :`{:(N_(1)V_(1),=,N_(2)V_(2)),("(Base)",,"(<a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>)"):}` <br/> `2.5xx(2)/(5)=(2)/(15)=(2)/(15)xxV_(2) "or" V_(2)=(15)/(2) <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> = 7.5 mL ` <br/>`BOH+HClrarrBCl+H_(2)O` <br/> 2.5 mL of 2 M base contain base<br/> `=2.5xx(2)/(5) = 1` mmol<br/> `:. ` Salt BCl formed = 1 mmol <br/> Volume of solution = 2.5 mL + 7.5 mL = 10 mL<br/> `:.` Conc. of salt [BCl] in the solution <br/> `=(1)/(10)M=0.1M` <br/> For salt of weak base and <a href="https://interviewquestions.tuteehub.com/tag/strong-653928" style="font-weight:bold;" target="_blank" title="Click to know more about STRONG">STRONG</a> acid <br/> `[H^(+)]=sqrt((K_(w)C)/(K_(b)))=sqrt((10^(-14)xx0.1)/(10^(-12)))=3.2xx10^(-2)M`</body></html>