What is the pH of a 0.30 M aqueous trisodium phosphate solution? K_a for monohydrogen phosphate ion is 4.2 × 10^-13 A)13.3 B)12.9 C)10.5 D)9.8 E)8.6

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Trisodium phosphate has the formula Na3PO4, which upon dissolution in water will dissociate into three sodium ions (Na+) and one polyatomic phosphate ion (PO43-). According to the given information, the acid dissociation constant (Ka) of the monohydrogen phosphate ion (HPO42-) is 4.2 x 10-13. This means that the stronger the acid, the smaller the Ka value, which indicates that HPO42- can act as a weak acid in water.
The equilibrium expression for the dissociation of HPO42- in water is:
HPO42-(aq) + H2O(l) ⇌ PO43-(aq) + H3O+(aq)
Since trisodium phosphate does not contain any H+ ions, we need to calculate the concentration of H3O+ ions produced from the dissociation of HPO42-. Let x be the concentration of H3O+ ions produced. Then, the concentration of PO43- ions produced will also be x, while the concentration of undissociated HPO42- ions will be 0.30 - x (assuming all of the HPO42- ions dissociate).
The Ka expression for HPO42- is:
Ka = [PO43-][H3O+]/[HPO42-] = (x)(x)/(0.30 - x)
Since Ka is a small value, we can assume that x is negligible compared to 0.30. Thus, we can simplify the expression as:
Ka = [PO43-][H3O+]/[HPO42-] ≈ (x)(x)/0.30
Rearranging and solving for x, we get:
x = sqrt(Ka[HPO42-]) = sqrt[(4.2 x 10-13)(0.30)] ≈ 7.3 x 10-7 M
Therefore, the pH of the solution can be calculated as:
pH = -log[H3O+] ≈ -log(7.3 x 10-7) ≈ 12.9
Thus, the answer is B) 12.9.

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