Question

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

Answer 1

<b>Note you had pH 3.39 in the problem.</b>
[H+] = 10^-3.30= 0.00501 M = [C6H4NO2-]
[HC6H4NO2] = 0.012 - 0.000501 =0.0115 M
<b>This step is OK.</b>

Ka = ( 0.00501)^2 / 0.0115 = 0.00218
pKa = - log Ka =2.66
<b>Note here that you used 0.00501 and not 0.000501 in the numerator. So the digits are ok but the power is off. I think it should be 2.18 x 10^-5 for Ka with the corresponding change in pKa. Check my work. </b>

Answer 2

What I got is a bit different... Can you please check my work?

[H+] = 10^-3.39 = 4.074*10^-4 = [C6H4NO2]
[HC6H4NO2] = 0.012-4.074*10^-4 = 0.0116 M

Ka = [4.074*10^-4]^2/0.0116 = 1.43*10^-5
pKa = -logKa = 4.84

Answer 3

good