At 25°C,the pH of a vinegar solution is 2.60.What are the values of [H₃O⁺] and [OH^-] in the solution? A)3.99 × 10^-12 M,2.51 × 10^-3 M B)2.51 × 10^-3 M,3.98 × 10^-12 M C)2.51 × 10^-3 M,11.40 M D)2.60 M,11.40 M

Question

Quizplus · Accepted Answer

Vinegar is an acidic solution of acetic acid (CH3COOH). In this case, the vinegar solution has a pH of 2.60 at 25°C. 
Using the definition of pH, we know that pH = -log[H+]. Therefore, [H+] = 10^(-pH). 
[H+] = 10^(-2.60) = 2.51 × 10^(-3) M.

Acetic acid is a weak acid, so it partially dissociates into its conjugate base (acetate) and a hydronium ion (H3O+). 
CH3COOH + H2O ⇌ CH3COO- + H3O+

The equilibrium constant expression for this reaction is Ka = [CH3COO-][H3O+]/[CH3COOH]. 
Since vinegar is a dilute solution, we can assume that the concentration of acetic acid before dissociation is equal to the initial concentration of vinegar solution.

At equilibrium, the amount of [CH3COO-] formed is equal to the amount of [H3O+] formed due to the 1:1 mole ratio. 
Therefore, [CH3COO-] = [H3O+] = x (let's assume this)

Using the expression for Ka and substituting the known values, we get: 
Ka = x^2 /[CH3COOH]
1.75 × 10^(-5) = x^2/ 2.51 × 10^(-3)
x = 1.49 × 10^(-3) M

Now, we can use the dissociation reaction of acetic acid to find the concentrations of the acetate and hydronium ions: 
[HSO2OH] = [CH3COOH] - x 
[HSO2OH] = 2.51 × 10^(-3) - 1.49 × 10^(-3) = 1.02 × 10^(-3) M

[HSO2O-] = [H3O+] = 1.49 × 10^(-3) M

[OHSO2-] = Kw/[HSO2OH] = (1.00 × 10^(-14))/ (1.02 × 10^(-3)) = 9.80 × 10^(-12) M

Therefore, the values of [HSO2OH] and [OHSO2-] in the vinegar solution are 1.02 × 10^(-3) M and 9.80 × 10^(-12) M, respectively.

The answer closest to these values is B: 2.51 × 10^(-3) M, 3.98 × 10^(-12) M.

At 25°C,the PH of a Vinegar Solution Is 2