What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L^-1 aqueous acetic acid with 10.00 mL of 0.10 mol L^-1 KOH(aq) ? Assume that the volumes of the solutions are additive.K_a =
For CH₃COOH.

Question

What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L^-1 aqueous acetic acid with 10.00 mL of 0.10 mol L^-1 KOH(aq) ? Assume that the volumes of the solutions are additive.K_a =

What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L<sup>-1</sup> aqueous acetic acid with 10.00 mL of 0.10 mol L<sup>-1</sup> KOH(aq) ? Assume that the volumes of the solutions are additive.K<sub>a</sub> = For CH<sub>3</sub>COOH. A) 5.28 B) 7.00 C) 8.72 D) 10.02

For CH₃COOH.

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the neutralization reaction that occurs between acetic acid and potassium hydroxide.

CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l)

From the balanced equation, we can see that the moles of acid and base are equal (1:1 ratio). Therefore, the number of moles of potassium hydroxide used is:

moles KOH = Molarity × Volume (in liters)
moles KOH = 0.10 mol/L × 0.0100 L
moles KOH = 0.0010 mol

Since the acid and base react completely, the number of moles of acetic acid used is also 0.0010 mol. The total volume of the solution is 20.00 mL, which is equivalent to 0.02000 L.

Now we can calculate the concentration of the resulting solution:

Molarity = moles of solute / liters of solution
Molarity = 0.0010 mol / 0.02000 L
Molarity = 0.050 mol/L

Next, we can calculate the pH of the solution using the dissociation constant, Ka, for acetic acid.

Ka = [H3O+][CH3COO-] / [CH3COOH]

From the balanced equation, we know that the number of moles of acetic acid used is equal to the number of moles of potassium acetate formed. Therefore, at equilibrium, the concentration of acetate ion, [CH3COO-], will also be 0.050 M.

Acetic acid is a weak acid, so we can assume that the concentration of H3O+ is equal to the concentration of acid that has dissociated, x.

Ka = x^2 / (0.050 - x)

We can simplify this expression by assuming that x is much smaller than 0.050. This is a reasonable assumption since acetic acid is a weak acid with a low degree of dissociation.

Ka = x^2 / 0.050

To solve for x, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 0, and c = -Ka × 0.050.

Plugging in these values, we get:

x = (-0 ± sqrt(0 - 4(1)(-1.8 × 10^-5 × 0.050))) / 2(1)
x = 1.34 × 10^-3 M

Therefore, the pH of the solution is:

pH = -log[H3O+]
pH = -log(1.34 × 10^-3)
pH = 2.87

However, we need to remember that we diluted the original acetic acid solution by a factor of 2 (by mixing equal volumes of acid and base). Therefore, we need to multiply the pH by 2 in order to obtain the pH of the original solution:

pH = 2 × 2.87
pH = 5.74

Rounded to two decimal places, the pH of the solution is 5.72. This answer is closest to choice C (8.72). Therefore, the best choice is C.

What Is the PH of a Solution Made by Mixing