Calculate the pH of the solution when 0.035 moles of OH- have been added to a 2.0 L containing equal amounts of 0.10 M acetic acid and 0.10 M sodium acetate. (pKa of acetic acid is 4.76)Group of answer choices4.834.895.084.91

To solve this problem, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this case, acetic acid is our HA and acetate is our A-.

Step 1: Calculate the initial moles of acetic acid and acetate. Since the molarity is 0.10 M for both and the volume of the solution is 2.0 L, the initial moles of both acetic acid and acetate are (0.10 mol/L) * 2.0 L = 0.20 mol.

Step 2: Calculate the moles of acetic acid and acetate after addition of OH-. When OH- is added, it will react with acetic acid to form water and acetate. So, the moles of acetic acid will decrease by 0.035 mol and the moles of acetate will increase by 0.035 mol. Therefore, the final moles of acetic acid and acetate are 0.20 mol - 0.035 mol = 0.165 mol and 0.20 mol + 0.035 mol = 0.235 mol, respectively.

Step 3: Calculate the concentrations of acetic acid and acetate after addition of OH-. The volume of the solution is still 2.0 L, so the concentrations are 0.165 mol / 2.0 L = 0.0825 M for acetic acid and 0.235 mol / 2.0 L = 0.1175 M for acetate.

Step 4: Substitute the values into the Henderson-Hasselbalch equation. pH = 4.76 + log(0.1175 / 0.0825) = 4.76 + 0.15 = 4.91

So, the pH of the solution after the addition of OH- is 4.91.