Asked by Saira
Question..
Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in the nicotinic acid has a pH of 3.39 at 25C. what is the acid-ionization constant, Ka and pKa for this acid at 25C?
Attempt..
HC6H4NO2 <----> C6H4NO2- + H+
[H+] = 10^-3.30= 0.00501 M = [C6H4NO2-]
[HC6H4NO2] = 0.012 - 0.000501 =0.0115 M
Ka = ( 0.00501)^2 / 0.0115 = 0.00218
pKa = - log Ka =2.66
Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in the nicotinic acid has a pH of 3.39 at 25C. what is the acid-ionization constant, Ka and pKa for this acid at 25C?
Attempt..
HC6H4NO2 <----> C6H4NO2- + H+
[H+] = 10^-3.30= 0.00501 M = [C6H4NO2-]
[HC6H4NO2] = 0.012 - 0.000501 =0.0115 M
Ka = ( 0.00501)^2 / 0.0115 = 0.00218
pKa = - log Ka =2.66
Answers
Answered by
DrBob222
See next question.
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