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
See next question.
Well, I must say, that seems like a pretty acidic situation! Maybe that's why they call it nicotinic "acid." But you know what they say, when life gives you acidic solutions, make some lemonade! Or in this case, calculate the Ka and pKa values.
It looks like you've got the right idea with the equation there. Now let's crunch those numbers. If the [H+] concentration is 10^-3.30, that means it's about as strong as a silent fart in a hurricane. And the concentration of [C6H4NO2-] is 0.00501 M, which makes it the star of the show.
Moving on to the acid-ionization constant, Ka. We take the square of the [H+] and divide it by the concentration of HC6H4NO2, which you calculated to be 0.0115 M. Mathematically speaking, that gives us a Ka value of 0.00218. Now, that's a pretty small Ka value, but hey, it's better than nothing!
And if you want to convert that Ka value into pKa, just take the negative logarithm of it. (Don't worry, I won't ask you to explain logarithms to me. I'd rather not get into a logarithmic nightmare.) So, after some number crunching, you get a pKa value of 2.66. Not too shabby!
So, to summarize, the acid-ionization constant, Ka, for nicotinic acid is 0.00218, and its pKa value is 2.66. Well done, my pH-balancing friend! Keep up the good work and remember, when life gets acidic, don't forget your sense of humor!
Your attempt is correct.
To find the acid-ionization constant (Ka), you can use the equilibrium expression for the dissociation of the acid:
Ka = [C6H4NO2-][H+] / [HC6H4NO2]
Given that [H+] = 10^-pH = 10^-3.39 = 0.00501 M and [C6H4NO2-] = [H+], the equation becomes:
Ka = (0.00501)^2 / 0.0115 = 0.00218
For pKa, you can take the negative logarithm of Ka:
pKa = -log Ka = -log 0.00218 = 2.66
So, the acid-ionization constant (Ka) for nicotinic acid is 0.00218 and the pKa is 2.66 at 25°C.
To determine the acid-ionization constant (Ka) and the pKa for the nicotinic acid (niacin), you need to use the pH and concentration information given in the question.
Step 1: Write the dissociation equation for the acid:
HC6H4NO2 <----> C6H4NO2- + H+
Step 2: Use the given pH to calculate the concentration of H+:
[H+] = 10^-pH
[H+] = 10^-3.39
[H+] = 0.00501 M
Step 3: Since nicotinic acid is a monoprotic acid, the concentration of [H+] is equal to the concentration of [C6H4NO2-].
Step 4: Determine the concentration of the acid [HC6H4NO2]:
[HC6H4NO2] = Initial concentration - concentration of [C6H4NO2-]
[HC6H4NO2] = 0.012 M - 0.00501 M
[HC6H4NO2] = 0.00699 M
Step 5: Use the concentrations of [C6H4NO2-] and [HC6H4NO2] to calculate the Ka:
Ka = ([C6H4NO2-] * [H+]) / [HC6H4NO2]
Ka = (0.00501 M * 0.00501 M) / 0.00699 M
Ka = 0.00218
Step 6: Calculate the pKa by taking the negative logarithm of the Ka value:
pKa = - log Ka
pKa = - log 0.00218
pKa ≈ 2.66
So, the acid-ionization constant (Ka) for nicotinic acid is approximately 0.00218, and the pKa is approximately 2.66 at 25°C.