1.

0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution . Calculate pH and hydroxyl ion concentration in the resulting solution assuming no change in the volume (k_(b) forpyridine = 1.5xx10^(-9)).

Answer» <html><body><p></p>Solution :Pyridine is a weak base. Thus, pyridine + pyridine chloride solution is a basic <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a>. Hence, <br/> `pOH = pK_(b) + log .(["Salt"])/(["Base"])` <br/> `pK_(b) = - log K_(b) = - log (1.5xx10^(-9))=9-0.1761=8.8239` <br/> [Pyridine ] = 0.23 M (Given ), [ Pyridinium chloride ] = `(0.15)/(500) xx <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> = 0.30 M` <br/> `:. pOH = 8.82+ log. (0.30)/(0.20) = 8.82 + 0.1761 = 8.896` <br/> i.e., `- log [OH^(-)]= 8.896or log [OH^(-)] = - 8.896 = <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>(9) . 104 or [OH^(-) ] = 1.271xx10^(-9)` <br/> `[OH^(-)] ` from `H_(2)O = 10^(-<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 cannot be neglected. <br/> Hence, total `[OH^(-)]=1.27xx10^(-9) + 10^(-7)= 10^(-9) (1.27+100) = 111.27 xx 10^(-9) M = 1.1127 xx 10^(-7) M` <br/> `[H^(+)] = (K_(w))/([OH^(-)])=(10^(-14))/(1.1127xx10^(-7))=8.987 xx 10^(-8)M` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = - log [H^(+)] = - log (8.987 xx 10^(-8))=8-0.9536= 7.0464`.</body></html>


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