Final answer:
Using the Henderson-Hasselbalch equation and given pKa value of 4.76 at pH 6.8, the ratio of the concentration of the deprotonated form to the protonated form of the acid is approximately 110:1.
Step-by-step explanation:
To find the ratio of the concentration of the protonated form to the deprotonated form of an acid in a solution at a particular pH, you can use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right)\)
Where \(\text{pH}\) is the pH of the solution, \(\text{pKa}\) is the acid dissociation constant, \([A^-]\) is the concentration of the deprotonated form, and \([HA]\) is the concentration of the protonated form. In this case, we know the \(\text{pKa}\) is 4.76 and the \(\text{pH}\) is 6.8. Let's rearrange the equation to solve for the ratio \(\frac{[A^-]}{[HA]}\):
\(\text{pH} - \text{pKa} = \log\left(\frac{[A^-]}{[HA]}\right)\)
\(6.8 - 4.76 = \log\left(\frac{[A^-]}{[HA]}\right)\)
\(2.04 = \log\left(\frac{[A^-]}{[HA]}\right)\)
To find the \(\frac{[A^-]}{[HA]}\) ratio, we calculate \(10^{2.04}\), which gives us approximately 110. Now we can say that the ratio of the deprotonated form to the protonated form of the acid in a pH environment of 6.8 is close to 110:1.