Respuesta :
Answer:
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
pH = 8.9
And this is the pH of the calcium phenylacetate solution
The calcium phenylacetate, Ca(C7H7COO)2, at 25 ºC pH is = 8.9
Calculation of Calcium phenylacetate
Then To translate this situation, After That, we need to note the overall reaction of the aqueous resolution of Ca(C₇H₇COO)₂:
Then Ca(C₇H₇COO)₂ <----------> Ca²⁺ + 2C₇H₇COO⁻
This define that the concentration of the phenylacetate is 2 times the concentration of the calcium phenylacetate:
That is [C₇H₇COO⁻] = 2 * 0.15 is = 0.30 M
After that We have the concentration of the phenylacetate, which is a conjugate base of the phenylacetic acid, Thus, we need the Kb of the conjugate. Now, This can be calculated using the subsequent declaration:
Kw is = Kb * Ka -----> Kb is = Kw/Ka
And also Ka is = 10^(-pKa)
Then We are Replacing the values (Accepting Kw is = 1x10⁻¹⁴)
Ka is = 10⁽⁻⁴°³²⁾ = 4.786x10⁻⁵
Kb is = 1x10⁻¹⁴ / 4.786x10⁻⁵ is = 2.089x10⁻¹⁰
Now that we have Kb, then we can compose the reaction of the phenylacetate in water, and also an ICE chart
Then C₇H₇COO⁻ + H₂O <--------> C₇H₇COOH + OH⁻ Kb is = 2.089x10⁻¹⁰
i) 0.30 0 0
e) 0.30-x x x
Kb is = [C₇H₇COOH] [OH⁻] / [C₇H₇COO⁻] Then Replacing the above values
2.089x10⁻¹⁰ is = x² / 0.30 - x
When Kb is very small, x then they would be a very small value too, so we can neglect 0.30 - x to 0.30 only:
2.089x10⁻¹⁰ is = x² / 0.30
2.089x10⁻¹⁰ * 0.30 is = x²
√6.27x10⁻¹¹ is = x
x is = 7.92x10⁻⁶ M
Now This is the concentration of the phenylacetate and also OH⁻. With this value then we can calculate the pOH and also then, the pH:
pOH is = -log[OH⁻]
pOH is = -log(7.92x10⁻⁶)
pOH is = 5.1
Finally the pH is:
pH is = 14 - pOH
pH is = 14 - 5.1
Therefore pH is = 8.9
So, This is the pH of the calcium phenylacetate solution
Find more information about Calcium phenylacetate here:
https://brainly.com/question/1392401
