Final answer:
To find the pH of the 0.348 M H2A solution with Ka1 of 4.38×10−7, we focus on the first dissociation and calculate the hydrogen ion concentration, which allows us to determine the pH value.
Step-by-step explanation:
To determine the pH of a 0.348 M H2A solution where H2A is a diprotic acid with a first dissociation constant (Ka1) of 4.38×10−7 and a second dissociation constant (Ka2) of 5.81×10−11, we primarily focus on the first dissociation as it contributes more significantly to the hydrogen ion concentration. The second dissociation has a much smaller effect on the pH and can often be ignored unless the pH is very high or the acid concentration is very low.
The reaction for the first dissociation is given by:
H2A(aq) ⇌ H+(aq) + HA−(aq)
Let the concentration of H+ ions produced be x. Then we can write the equilibrium expression for the first dissociation:
Ka1 = [H+][HA−]/[H2A]
Assuming x is small compared to the initial concentration and can be ignored in the denominator, we get:
Ka1 ≈ x2 / (0.348 - x) ≈ x2 / 0.348
Solving for x gives us:
x = √(Ka1 × 0.348) = √(4.38×10−7 × 0.348)
After calculating x, we obtain the hydrogen ion concentration and can then calculate the pH as follows:
pH = -log[H+] = -log(x)