Respuesta :
Answer : The pH of a solution is, 4.50
Explanation : Given,
[tex]K_a=6.30\times 10^{-5}[/tex]
Concentration of benzoic acid (Acid) = 0.150 M
Concentration of sodium benzoate (salt) = 0.300 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (6.30\times 10^{-5})[/tex]
[tex]pK_a=5-\log (6.30)[/tex]
[tex]pK_a=4.20[/tex]
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=4.20+\log (\frac{0.300}{0.150})[/tex]
[tex]pH=4.50[/tex]
Thus, the pH of a solution is, 4.50
The pH of the solution prepared by benzoic acid and sodium benzoate has been 4.5.
The [tex]\rm \bold{pK_a}[/tex] of the solution has been calculated based on the [tex]\rm \bold{k_a}[/tex] value.
[tex]\rm \bold{pK_a}[/tex] = -log [tex]\rm \bold{k_a}[/tex]
Given, [tex]\rm \bold{k_a}[/tex] = 6.30 [tex]\rm \times\;10^-^5[/tex]
[tex]\rm \bold{pK_a}[/tex] = -log (6.30 [tex]\rm \times\;10^-^5[/tex])
[tex]\rm \bold{pK_a}[/tex] = 4.20
The pH of the solution can be given by:
pH = [tex]\rm pk_a\;+\;log\;\dfrac{[salt]}{[acid]}[/tex]
The salt has been the sodium benzoate with a concentration of 0.30 mol.
The concentration of acid (benzoic acid) has been 0.150 mol.
Substituting the values:
pH = 4.20 + [tex]\rm log\;\dfrac{0.3}{0.15}[/tex]
pH = 4.20 + log 2
pH = 4.20 + 0.30
pH = 4.5
The pH of the solution prepared by benzoic acid and sodium benzoate has been 4.5.
For more information about the pH of the solution, refer to the link:
https://brainly.com/question/1265333
