what is the composition of 0.05M potassium phthalate buffer ,pH 5.00?
phathalic acid (PH2,PH^-1,P^-2)is dibasic with pKa values of 2.95 and 5.41
To find the composition of a 0.05M potassium phthalate buffer at pH 5.00, we need to use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa values of the acid and its conjugate base:
pH = pKa + log ([A^-] / [HA])
In this case, the acid is phthalic acid (HA) and its conjugate base is phthalate ion (A^-). The pKa values given are 2.95 and 5.41.
At the pH of 5.00, we know that the pH is in between the pKa values. To determine the composition, we need to calculate the ratio of [A^-] / [HA] using the Henderson-Hasselbalch equation.
Let's start by rearranging the equation to solve for the ratio [A^-] / [HA]:
[A^-] / [HA] = 10^(pH - pKa)
[A^-] / [HA] = 10^(5.00 - 5.41)
[A^-] / [HA] = 10^(-0.41)
[A^-] / [HA] = 0.394
This means that the ratio of phthalate ion to phthalic acid in the buffer is 0.394. Since the concentration of the buffer is given as 0.05M, we can calculate the actual concentrations:
[HA] = (0.394 / (0.394 + 1)) * 0.05M
[HA] = 0.394 / 1.394 * 0.05M
[HA] = 0.0142M
[A^-] = (1 / (0.394 + 1)) * 0.05M
[A^-] = 1 / 1.394 * 0.05M
[A^-] = 0.0358M
Therefore, the composition of the 0.05M potassium phthalate buffer at pH 5.00 is approximately 0.0142M phthalic acid (HA) and 0.0358M phthalate ion (A^-).