A 40.00 mL sample of 0.1000 M diprotic acid (H2A) is titrated with 0.0900 M KOH. What
volume KOH must be added to give a pH of 5.697? Ka1 = 1.42 × 10−3
and Ka2 = 2.01 × 10−6
.
(MW KOH = 56.01 g/mol)
To solve this problem, we need to first determine the concentration of H2A at the start of the titration. We can use the formula for diprotic acid:
[H2A] = [H2A-initial] - [H+]
Since H2A is a diprotic acid, it can donate two protons. We know the initial concentration of H2A is 0.1000 M, so the concentration of H2A after donating one proton is (0.1000 - x), where x is the concentration of H+.
Since the pH is given as 5.697, we can calculate the concentration of H+ using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-5.697) = 2.89 * 10^(-6) M
Therefore, the concentration of H2A after donating one proton is:
0.1000 M - 2.89 * 10^(-6) M = 0.09999711 M
Now, we can calculate the concentration of OH- (hydroxide ions) that reacts with H2A. Since H2A is a diprotic acid, it reacts with two OH- ions to form water. Therefore, the concentration of OH- ions is twice the concentration of H2A after donating one proton:
[OH-] = [H2A after donating one proton] * 2
[OH-] = 0.09999711 M * 2 = 0.1999942 M
To determine the volume of the KOH solution needed to reach the desired pH, we can use the equation:
[H+] * [OH-] = Kw
where Kw is the ion product of water and is equal to 1.0 x 10^(-14) at 25 degrees Celsius.
(2.89 * 10^(-6) M) * (0.1999942 M) = 1.0 x 10^(-14) M^2
Let's call the volume of KOH solution needed V.
V * 0.0900 M = 0.1999942 M * 40.00 mL
V = (0.1999942 M * 40.00 mL) / (0.0900 M)
V = 88.874 mL
Therefore, approximately 88.874 mL of the 0.0900 M KOH solution should be added to the 40.00 mL sample of 0.1000 M diprotic acid to reach a pH of 5.697.