A solution of nicotinic acid, HC6H4NO2, of concentration 0.012 M has a pH of 3.39 at 25 degrees Celsius. Calculate Ka for this acid.
To calculate the Ka for the nicotinic acid, we need to use the equation for the dissociation of an acid:
HC6H4NO2 (aq) ⇌ H+ (aq) + C6H4NO2- (aq)
Nicotinic acid is a weak acid, so we can assume that the amount of dissociated acid is much less than the initial concentration of the acid. Therefore, we can use the assumption that the change in concentration of the acid is negligible compared to the initial concentration.
The pH of a solution is related to the concentration of H+ ions present. The pH is defined as the negative logarithm of the H+ concentration.
pH = -log[H+]
From the given pH, we can calculate the H+ concentration using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-3.39) = 4.35 × 10^(-4) M
Since the acid is weak, we can assume that the concentration of the conjugate base (C6H4NO2-) is equal to the H+ concentration.
Now, we can set up the equation for the dissociation of the acid and write the expression for the equilibrium constant Ka:
Ka = [H+] * [C6H4NO2-] / [HC6H4NO2]
Since the concentration of the acid is given as 0.012 M, we have:
Ka = (4.35 × 10^(-4)) * (4.35 × 10^(-4)) / (0.012)
= 0.156
Therefore, the value of Ka for nicotinic acid is approximately 0.156.