Posted by Rushi on Sunday, March 15, 2009 at 3:35pm.

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

Responses

* Plz -- Chem Help - DrBob222, Sunday, March 15, 2009 at 5:47pm

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

Ka = ( 0.00501)^2 / 0.0115 = 0.00218
pKa = - log Ka =2.66
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.

* Plz -- Chem Help - Ed, Monday, March 16, 2009 at 11:22pm

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 1

This looks ok to me now. The discrepancy before was two fold.
You had 3.39 in the problem but used 3.30 AND you did not use the correct power of 10 when you substituted into the Ka expression.

Answer 2

I looked in The Merck Index and it lists the pKa for nicotinic acid as 4.85.

Answer 3

hey guys,

the reason why you have off answers is because you forgot to make the
assumption tthat if the ionization of nitotinic acid is less than 5% than the 0.012M - 4.07*10^-4=0.012M

to check the assumption divide the 4.07*10^-4/0.012M and multiply by 100 to see if the number comes less den 5% or over it..

If it is less den assumption is good otherwise you need to do the substraction in the denominator.

and in this question assumption is good so no need to do that substration.

thanks for reading my reply.