Solution:

Given,
Volume of CH3COOH solution = 500 mL
Concentration of CH3COOH solution = 0.1 M
Volume of NaOH solution = 250 mL
Concentration of NaOH solution = 0.1 M
pKa of acetic acid = 4.74

Step 1: Write the balanced chemical equation for the reaction between CH3COOH and NaOH.

CH3COOH + NaOH → NaCH3COO + H2O

Step 2: Calculate the number of moles of CH3COOH and NaOH.

Number of moles of CH3COOH = Molarity × Volume (in liters) = 0.1 × 0.5 = 0.05 moles
Number of moles of NaOH = Molarity × Volume (in liters) = 0.1 × 0.25 = 0.025 moles

Step 3: Determine the limiting reagent.

Since NaOH is present in a lesser amount than CH3COOH, it is the limiting reagent.

Step 4: Calculate the number of moles of NaCH3COO formed.

From the balanced chemical equation, the stoichiometric ratio between NaOH and NaCH3COO is 1:1. Therefore, the number of moles of NaCH3COO formed is also 0.025 moles.

Step 5: Calculate the concentration of the resulting solution.

Total volume of the resulting solution = Volume of CH3COOH solution + Volume of NaOH solution = 500 mL + 250 mL = 750 mL = 0.75 L

Concentration of NaCH3COO = Number of moles of NaCH3COO / Total volume of the resulting solution = 0.025 / 0.75 = 0.033 M

Step 6: Calculate the pH of the resulting solution.

Using the Henderson-Hasselbalch equation,

pH = pKa + log ([A-]/[HA])

where [A-] is the concentration of NaCH3COO and [HA] is the concentration of CH3COOH.

Substituting the values,

pH = 4.74 + log (0.033/0.05)

pH = 4.74 - 0.23

pH = 4.51

Therefore, the pH of the resulting solution is 4.51.

Conclusion:

The pH of the mixture of 500 mL 0.1 M CH3COOH and 250 mL 0.1 M NaOH solution is 4.51.

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