Final answer:
The pH of a 0.227 M H3PO4 solution is found by using the first dissociation constant, as it's the primary contributor to pH. By simplifying the expression due to the small size of Ka1 and disregarding x in the denominator, we calculate the pH to be approximately 1.42, making the correct answer choice (d).
Step-by-step explanation:
To calculate the pH of the 0.227 M H3PO4 solution, we must consider the dissociation constants (Ka) given for the acid. H3PO4 is a triprotic acid, meaning it donates three protons (H+), each governed by its own Ka. However, because Ka1 >> Ka2 >> Ka3, the first dissociation will be the primary contributor to the pH of the solution.
We can use the first dissociation constant (Ka1 = 7.5 × 10−3) and the concentration of H3PO4 to set up an ICE table (Initial, Change, Equilibrium). Assuming x is the concentration of H+ ions at equilibrium, we can write:
H3PO4 → H+ + H2PO4−
Initial (M): 0.227, 0, 0
Change (M): −x, +x, +x
Equilibrium (M): 0.227−x, x, x
The equilibrium expression is:
Ka1 = [H+][H2PO4−] / [H3PO4]
Inserting the values, we have:
7.5 × 10−3 = x2 / (0.227−x)
Because Ka1 is relatively small, we can assume that x is much smaller than 0.227 and therefore, ignore x in the denominator. This simplifies the expression to:
7.5 × 10−3 × 0.227 = x2
The resulting x is the concentration of H+ in the solution. From this, we calculate the pH, which is the negative logarithm of the H+ concentration (pH = −log[H+]).
Performing the calculations, we find that the pH is approximately 1.42, which is answer choice (d).