0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)

0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2

1.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)
Answer» <html><body><p><br/></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> :Pyridiniumchloride = 0.15 mole <br/>Pyridine <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> ` cm ^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)"of "0.2 M = 0.1 ` mole<br/>` pOH = pK_b +log "" ([s]) /(["Base" ]) ` <br/>` =- log (1.5 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) ) +log"" (0.15)/(0.1) ` <br/> ` =9- log 1. 5+ log 1.5 =9` <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- 9 = 5`</body></html>