What is the pH of the resulting solution if 25.00 mL of 0.10 mol L^-1 acetic acid is added to 10.00 mL of 0.10 mol L^-1 NaOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH A) 9. 43 B) 9.08 C) 4.92 D) 4. 57

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of moles of acetic acid and NaOH added:
moles of acetic acid = 0.1 mol/L x (25.00 mL/1000 mL) = 0.0025 mol
moles of NaOH = 0.1 mol/L x (10.00 mL/1000 mL) = 0.0010 mol

Since acetic acid and NaOH react in a 1:1 ratio, all the moles of NaOH will be used up by the acetic acid. This means that the remaining moles of acetic acid will be in solution and will determine the pH of the solution.

The initial moles of acetic acid in solution would be:
moles of acetic acid = 0.1 mol/L x (35.00 mL/1000 mL) = 0.0035 mol

Since acetic acid is a weak acid, we need to use the dissociation constant (Ka) to calculate the pH of the solution. The equation for the dissociation of acetic acid is as follows:
CH3COOH + H2O ⇌ CH3COO- + H3O+

The Ka for acetic acid is 1.8 x 10^-5. We can set up an ICE table to help us calculate the pH:

I: 0.0035 mol/L (initial concentration of acid)
C: -x mol/L (concentration change)
E: (0.0035-x) mol/L (equilibrium concentration of acid)

Ka = [CH3COO-][H3O+] / [CH3COOH]
1.8 x 10^-5 = x^2 / (0.0035 - x)

Solving the quadratic equation, we get x = 6.30 x 10^-4
pH = -log[H3O+] = -log(6.30 x 10^-4) = 3.20

However, we need to remember that we calculated the pH for the remaining acetic acid only, not for the entire solution. The NaOH that was added has contributed additional OH- ions to the solution. So we need to calculate the concentration of OH- ions before we can calculate the final pH.

The amount of NaOH added is 0.1 mol/L x (10.00 mL/1000 mL) = 0.0010 mol. Since NaOH is a strong base, it completely dissociates in solution. This means that the concentration of OH- ions in solution is equal to the amount of NaOH added, divided by the total volume of the solution:

[OH-] = 0.0010 mol / (25.00 mL + 10.00 mL) / 1000 mL/mL = 0.033 mol/L

To calculate the final pH, we can use the following equation:
pH = 14 - pOH = 14 - (-log[OH-]) = 14 - (-log0.033) = 14 + 1.48 = 15.48

Therefore, the pH of the resulting solution is 15.48, which corresponds to answer choice D.

What Is the PH of the Resulting Solution If 25