Formic acid (CH2O2) is used as an antibacterial on livestock feed. A sample of formic acid has been found to have a pH = 3, and data shows it has a Ka = 1.6E-4 M. What is the concentration of the sample?
This question tests your knowledge on weak acids, pH, Ka and equilibrium. (Note: E notation has been used, E-3 is x10-3 ). Formic acid is a weak acid and so does not fully dissociate, the degree of dissociation is measured by the Ka. Ka = [A-}{H+}/[AH]. Using pH=-log[H+] we find that [H+] is 0.001M. The concentration of protons is the same as the conjugate base so, [H+] = [A-]. We can now rearrange the Ka equation to give us the concentration of the acid [AH] = [H+]2 / Ka. [AH] = {1E-3M]2/ 1.6E-4M. [AH] = 6.25E-3M. We should report our result to the same amount of decimal places that the question gives us so: the sample has a concentration of 6.3E-3M.
