A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine.

A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate

1.	A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine.
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> :Pyridinium hydrochloride `C_6H_5N^(+) <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>^(-)` is conjugate <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> of pyridine `(C_6H_5N)`. <br/> pH=3.44 = -log `[H^+]` <br/> `therefore` log `[H^+] = -3.44 = bar4.56` <br/> `therefore [H^+] = 10^(-3.44) = 3.63xx10^(-4) = alpha` <br/> `{:(,C_6H_5N^(+)HCl^(-)+ aqhArr ,C_6H_5NCl_((aq))^(-) +, H_((aq))^(+)),("Molarity at equilibrium" , (0.012-alpha)M =0.02 M, 3.63xx10^(-4) , 3.63xx10^(-4)) :}` <br/> `K_a=([C_6H_5NCl^(-)][H^+])/([C_6H_5N^(+)HCl^-])` <br/> `=((3.63xx10^(-4))(3.63xx10^(-4)))/0.02=6.588xx10^(-6)` <br/> Pyridine is conjugate base of pyridinium hydrochloride , so its ionization constant = `K_b` <br/> `K_b=K_w/K_a=(1.0xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>))/(6.588xx10^(-6))=1.5179xx10^(-7)`</body></html>