Final answer:
The pH of the solution after adding 29.0 mL of NaOH in the titration can be calculated using the Henderson-Hasselbalch equation. By considering the stoichiometry and reaction between CH3COOH and NaOH, we find that the pH is 4.88.
Step-by-step explanation:
In this titration, a 52.0-mL volume of 0.350 mol L−1 CH3COOH (with a Ka value of 1.8×10−5) is titrated with 0.400 mol L−1 NaOH. To calculate the pH after adding 29.0 mL of NaOH, we need to consider the stoichiometry and reaction between CH3COOH and NaOH.
First, calculate the amount of moles of CH3COOH initially present:
0.0520 L × 0.350 mol L−1 = 0.0182 mol CH3COOH
Next, consider the reaction:
CH3COOH + NaOH -> CH3COONa + H2O
Since CH3COOH and NaOH react in a 1:1 stoichiometric ratio, after adding the 29.0 mL of NaOH, the amount of moles of CH3COOH remaining is:
0.0182 mol - (0.0290 L × 0.400 mol L−1) = 0.00794 mol CH3COOH
Now, we can calculate the moles of CH3COONa formed:
0.0290 L × 0.400 mol L−1 = 0.0116 mol CH3COONa
Since CH3COOH and CH3COONa are a conjugate acid-base pair, the pH of the solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where [A-] represents the concentration of the conjugate base (CH3COONa) and [HA] represents the concentration of the weak acid (CH3COOH) remaining in the solution.
Using the given pKa value of 1.8×10−5:
pH = -log(1.8×10−5) + log(0.0116/0.00794)
pH = 4.74 + 0.14
pH = 4.88