Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in nicotinic acid has a pH of 3.39 at 25*C. What is the acid-ionization constant, Ka and pKa for this acid at 25*C?
Write the ionization equation.
Prepare an ICE chart.
You know pH and this will give you (H^+), (nicotine ion), and nicotinic acid concn. Plug these into the Ka expression and solve for Ka, then pKa. Post your work if you get stuck.
To find the acid-ionization constant, Ka, and pKa for nicotinic acid (niacin), we need to make use of the pH and concentration of the acid solution.
First, let's write down the balanced equation for the ionization of the nicotinic acid:
HC6H4NO2 ⇌ H+ + C6H4NO2-
The Ka expression for this reaction can be written as:
Ka = [H+][C6H4NO2-] / [HC6H4NO2]
Given that the solution has a pH of 3.39, we can determine the concentration of H+ ions using the equation:
pH = -log[H+]
3.39 = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-3.39)
Now, let's consider the acid concentration provided in the question, which is 0.012 M. This is the concentration of HC6H4NO2.
Substituting the values into the Ka expression:
Ka = [H+][C6H4NO2-] / [HC6H4NO2]
= (10^(-3.39))[C6H4NO2-] / 0.012
We can now solve for the concentration of C6H4NO2-:
0.012 = (10^(-3.39))[C6H4NO2-]
[C6H4NO2-] = 0.012 / (10^(-3.39))
Once we have the concentration of C6H4NO2-, we can substitute it back into the Ka expression to solve for Ka:
Ka = (10^(-3.39))(0.012 / (10^(-3.39)))
Finally, to find pKa, we take the negative logarithm of Ka:
pKa = -log(Ka)
By substituting the values, you will be able to calculate the acid-ionization constant (Ka) and pKa for nicotinic acid at 25°C.