"Phenylacetic acid, C7H7COOH, builds up in the blood of people with phenylketonuria. The pKa of phenylacetic acid is 4.32 at 25 ºC. What is the pH of a 0.15 M aqueous solution of calcium phenylacetate, Ca(C7H7COO)2, at 25 ºC?"

pH = 8.9

Explanation:

To solve this problem, we need to write the overall reaction of the aqueous solution of Ca(C₇H₇COO)₂:

Ca(C₇H₇COO)₂ <> Ca²⁺ + 2C₇H₇COO⁻

This means that the concentration of the phenyl acetate is 2 times the concentration of the calcium phenylacetate:

[C₇H₇COO⁻] = 2 * 0.15 = 0.30 M

We have the concentration of the phenylacetate, this is a conjugate base of the phenylacetic acid, therefore, we need the Kb of the conjugate. This can be calculated using the following expression:

Kw = Kb * Ka > Kb = Kw/Ka

And Ka = 10^(-pKa)

Replacing the values (Assuming Kw = 1x10⁻¹⁴)

Ka = 10⁽⁻⁴°³²⁾ = 4.786x10⁻⁵

Kb = 1x10⁻¹⁴ / 4.786x10⁻⁵ = 2.089x10⁻¹⁰

Now that we have Kb, we can write the reaction of the phenylacetate in water, and an ICE chart

        C₇H₇COO⁻ + H₂O <> C₇H₇COOH + OH⁻    Kb = 2.089x10⁻¹⁰

i)           0.30                                        0                0

e)          0.30-x                                     x                 x

Kb = [C₇H₇COOH] [OH⁻] / [C₇H₇COO⁻]      Replacing the above values

2.089x10⁻¹⁰ = x² / 0.30 - x    

As Kb is very small, x would be a very small value too, so we can neglect 0.30 - x to 0.30 only:

2.089x10⁻¹⁰ = x² / 0.30

2.089x10⁻¹⁰ * 0.30 = x²

√6.27x10⁻¹¹ = x

x = 7.92x10⁻⁶ M

This is the concentration of the phenylacetate and OH⁻. With this value we can calculate the pOH and then, the pH:

pOH = -log[OH⁻]

pOH = -log(7.92x10⁻⁶)

pOH = 5.1

Finally the pH:

pH = 14 - pOH

pH = 14 - 5.1

i believe the answer is a.

wher is the rest of the question