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0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and the solution diluted to one litre. The H^(+) ion concentration in the solution will be

Answer» <html><body><p>`1.6xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)`<br/>`8xx10^(-11)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5xx10-1901302" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX10">5XX10</a>^(-5)`<br/>`8xx10^(-2)`</p>Solution :`{:(,CH_(3)NH_(2),+,HCl,rarr,CH_(3)N overset(+)H_(3)Cl^(-),,,),("Initial conc.",0.1,,0.08,,0 "mol" L^(-1),,,),("Conc. in solution after mixing",0.1-0.08 ,,0,,0.08 "mol " L^(-1),,,):}` <br/> `=0.02` (HCl being limiting <a href="https://interviewquestions.tuteehub.com/tag/reactant-1178091" style="font-weight:bold;" target="_blank" title="Click to know more about REACTANT">REACTANT</a>) <br/> Thus, the final solution is abasic buffer of `CH_(3)NH_(2) and CH_(3)NH_(3)^(+)Cl` <br/> `:. pOH = pK_(b) + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>. (["Salt"])/(["Base"])`<br/> i.e., `-log [OH^(-)]=-log K_(b) + log. (["Salt"])/(["Base"])` <br/> or `[OH^(-)]=K_(b) (["Salt"])/(["Base"]) = 5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)xx(0.02)/(0.08) ` <br/> `=(5)/(4) xx 10^(-4)M = 1.25 xx 10^(-4)M` <br/> `:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)`</body></html>


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