A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol of ammonium chloride to make a buffer solution of 1 L. Calculate the pH of the resulting buffer solution. Dissociation constant of ammonium hydroxide at 25^(@)C is 1.81xx10^(-5).

A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol

1.	A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol of ammonium chloride to make a buffer solution of 1 L. Calculate the pH of the resulting buffer solution. Dissociation constant of ammonium hydroxide at 25^(@)C is 1.81xx10^(-5).
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`pOH = pK_(b) + log. (["Salt"])/(["Base"])=pK_(b) + log. ([NH_(4)Cl])/([NH_(4)OH])` <br/>`[NH_(4)OH]=0.40` mol `L^(-1)` <br/> `[NH_(4)Cl]=0.50 ` mol `L^(-1)` <br/> `pK_(b) = - log K_(b) = - log (1.81 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))=5-0.2577 = 4.7423` <br/> `:. pOH = 4.742 + log. (0.5)/(0.4)= 4.742+ log 1.25 = 4.742 + 0.0969 = 4.8389 ~= 4.839` <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> = 14-pOH = 14 - 4.839 = 9.161`.</body></html>