Answer:
pH = 3.74
Explanation:
<u>Given:</u>
Initial volume of CH3COOH, V1 = 25.00 ml
Initial concentration of CH3COOH, M1 = 0.10 M
Initial volume of CH3COONa, V1 = 25.00 ml
Initial concentration of CH3COONa, M2 = 0.010 M
Ka (CH3COOH) = 1.8*10^-5
<u>To determine:</u>
pH of the solution
<u>Calculation:</u>
The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:
![pH = pKa + log\frac{[A-]}{[HA]} ----(1)](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%20----%281%29)
where A- = concentration of conjugate base = [CH3COONa]
HA = weak acid = [CH3COOH]
Step 1: Calculate the final concentration of CH3COONa
V1 = 25.00 ml
V(final) = Total volume = 25.00 + 25.00 = 50.00 ml
M1 = 0.010 M

<u>Step 2</u>: Calculate the final concentration of CH3COOH
V1 = 25.00 ml
V(final) = Total volume = 25.00 + 25.00 = 50.00 ml
M1 = 0.10 M

<u>Step 3</u>: Calculate the pH
Based on equation (1)
![pH = pKa + log\frac{[CH3COONa]}{[CH3COOH]} ----(1)](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5BCH3COONa%5D%7D%7B%5BCH3COOH%5D%7D%20----%281%29)
pKa = -log Ka = -log(1.8*10^-5) = 4.74
![pH = -logKa + log\frac{[CH3COONa]}{[CH3COOH]}](https://tex.z-dn.net/?f=pH%20%3D%20-logKa%20%2B%20log%5Cfrac%7B%5BCH3COONa%5D%7D%7B%5BCH3COOH%5D%7D)
![pH = 4.74 + log\frac{[0.005]}{[0.05]}](https://tex.z-dn.net/?f=pH%20%3D%204.74%20%2B%20log%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.05%5D%7D%20)
pH = 3.74