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
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.
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
good
To calculate the acid-ionization constant, Ka, and pKa for the nicotinic acid (niacin) solution, you first need to understand the relationship between pH and the concentration of the acid and its conjugate base.
1. Start with the balanced equation for the ionization of nicotinic acid:
HC6H4NO2 ⇌ C6H4NO2- + H+
2. Use the pH value given to determine the concentration of [H+]. Since pH is defined as the negative logarithm of [H+], you can calculate [H+] using the equation:
[H+] = 10^(-pH)
In this case, [H+] = 10^(-3.39) = 0.00499 M, which can be approximated as 0.005 M.
3. Since nicotinic acid is a monoprotic acid, the concentration of [C6H4NO2-] will be equal to the concentration of [H+] based on the stoichiometry of the reaction.
[C6H4NO2-] = [H+] = 0.005 M
4. Subtract the concentration of [H+] from the initial concentration of nicotinic acid, [HC6H4NO2], to find its concentration:
[HC6H4NO2] = initial concentration - [H+]
[HC6H4NO2] = 0.012 M - 0.005 M
[HC6H4NO2] = 0.007 M
5. Now that you have the concentrations of [H+], [C6H4NO2-], and [HC6H4NO2], you can calculate the acid-ionization constant, Ka, using the equation:
Ka = ([C6H4NO2-] * [H+]) / [HC6H4NO2]
Ka = (0.005 M * 0.005 M) / 0.007 M
Ka = 0.002857 M
6. Finally, calculate pKa by taking the negative logarithm of Ka:
pKa = -log(Ka)
pKa = -log(0.002857)
pKa = 2.55
Therefore, the acid-ionization constant (Ka) for nicotinic acid is 0.002857 and the pKa is 2.55.