Determine the [H30+] of a 0.120 M solution of benzoic acid. Ph?

Question

How can I determine the concentration of hydrogen ions, [H₃O⁺], in a 0.120 M solution of benzoic acid? Additionally, what would be the pH of this solution?

Anonymous · Accepted Answer

Benzoic acid is a weak acid and therefore it does not dissociate completely. To do this question properly you must be given, or be able to obtain, data about the degree of dissociation of benzoic acid:

pKa for C6H5COOH = 4.19

Ka = 10^-4.19

Ka = 6.456*10^-5

Ka = [H+] * [ C6H5COO-]/[C6H5COOH]

Because benzoic acid is monoprotic, [H+] = [C6H5COO-]

And because the dissociation of H+ is very small, we can use the molar concentration of the acid without allowing for the quantity dissociated:

Therfore we can write:

Ka = [H+]²/[C6H5COOH]

Substituting:

6.456*10^-5 = [H+]² / 0.120

[H+] = √(6.45610^-5) 0.120

[H+] = √(7.75*10^-6

[H+] = 2.78*10^-3

pH = -log H+

pH = -log 2.78*10^-3

pH = 2.56

Answers: [H3O+] = 2.78*10^-3M

pH = 2.56

Bria Dietrich · Answer

The acid dissociation constant (Ka) for benzoic acid is 6.3x10^-5
Let HA represent benzoic acid (C6H5COOH).
HA <--> H^+1 + A^-1
Ka = [H+][A-]/[HA] = 6.3x10^-5
[H+] must equal [A-]
let X = [H+]
Ka = 6.3x10^-5 = X^2 / (0.120 M - X)
From the value of Ka you can assume that [H+] is much smaller than 0.120 M so you can assume 0.120 M - X = 0.120 M 
6.3x10^-5 = X^2 / 0.120
X = [H+] = 0.00275 mol/L and pH = 2.56
Without the assumption you would need to solve the quadratic equation for X (the value will be close to 0.00272
Since you only had 2 sig fiq for Ka, report the pH as 2.6

Anonymous · Answer

Benzoic acid, C6H5CO2H, is a vulnerable acid (Ka = 6.3 x 10-5). Calculate the preliminary concentration (in M) of benzoic acid it particularly is had to offer an aqueous answer of benzoic acid that has a pH of two.fifty 4.

Anonymous · Answer

You need a Ka value so u can set up an ICE table since benzoic acid is a weak acid and does not dissociate 100%

Adan Nicolas I · Answer

benzoic acid is monoprotic, so [H3O+] would be 0.12M.
using pH = -log[H+], pH = 0.92.
Edit: obviously it's not a strong acid, but when you get a question like this on a test for example, you assume it ionises 100%.

Anonymous

Determine the [H30+] of a 0.120 M solution of benzoic acid. Ph?

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