A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8xx10^(-5). What is the mole proportion of ammonium chloride and ammonium hydroxide?

A buffer solution of pH 8.3 is prepared from ammonium chloride and amm

1.	A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8xx10^(-5). What is the mole proportion of ammonium chloride and ammonium hydroxide?
Answer» <html><body><p></p>Solution :pOH = 14 - pH = 14 - 8.3 = 5.7 <br/> `pH=pK_(a)+"log"(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Base"])rArrpH=-log_(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)K_(a)+"log"(["Salt"])/(["Base"])` <br/> `5.7=-log_(10)(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.8xx10^(-5))+"log"([NH_(4)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>])/([NH_(4)OH])rArr5.7=5-log1.8+"log"([NH_(4)Cl])/([NH_(4)OH])` <br/> `5.7=5-0.2553+"log"([NH_(4)Cl])/([NH_(4)OH])rArr5.7=4.7447+"log"([NH_(4)Cl])/([NH_(4)OH])` <br/> `therefore([NH_(4)Cl])/([NH_(4)OH])=5.7-4.7447=0.9553` <br/> `rArr([NH_(4)Cl])/([NH_(4)OH])="anti"log(0.9553)=9.022""` The ratio is 9 : 1.</body></html>