Respuesta :
Answer : The pH of a solution is, 4.5
Explanation : Given,
Moles of benzoic acid = 0.15 mol
Moles of sodium benzoate = 0.30 mol
Volume of solution = 1.00 L
The dissociation constant for benzoic acid = [tex]K_a=6.5\times 10^{-5}[/tex]
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 (6.5\times 10^{-5})[/tex]
[tex]pK_a=5-\log (6.5)[/tex]
[tex]pK_a=4.2[/tex]
Now we have to calculate the pH of a solution.
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=4.2+\log [\frac{(\frac{0.30}{1.00L})}{(\frac{0.15}{1.00L})}][/tex]
[tex]pH=4.5[/tex]
Therefore, the pH of a solution is, 4.5
The pH of a solution prepared by dissolving 0.15 mol benzoic acid is 4.5.
Calculation of pH of a solution:
Since
Moles of benzoic acid = 0.15 mol
Moles of sodium benzoate = 0.30 mol
Volume of solution = 1.00 L
And, dissociation constant for benzoic acid is 6.50 x 10-5
So, here pKa should be
= -log(6.5*10^-5)
= 5-log(6.5)
= 4.2
Now the pH should be
= 4.2 + log(0.30/1.00L/0.15/1.00L)
= 4.5
Hence, The pH of a solution prepared by dissolving 0.15 mol benzoic acid is 4.5.
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