<u>Answer:</u> The pH of the solution is 4.14
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
Molarity of NaOH = 1 M
Volume of solution = 20 mL
Putting values in above equation, we get:
![1M=\frac{\text{Moles of NaOH}\times 1000}{20mL}\\\\\text{Moles of NaOH}=\frac{1\times 20}{1000}=0.02mol](https://tex.z-dn.net/?f=1M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20NaOH%7D%5Ctimes%201000%7D%7B20mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20NaOH%7D%3D%5Cfrac%7B1%5Ctimes%2020%7D%7B1000%7D%3D0.02mol)
Molarity of acetic acid solution = 0.26 M
Volume of solution = 100 mL
Putting values in above equation, we get:
![1M=\frac{\text{Moles of acetic acid}\times 1000}{100mL}\\\\\text{Moles of acetic acid}=\frac{1\times 100}{1000}=0.1mol](https://tex.z-dn.net/?f=1M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20acetic%20acid%7D%5Ctimes%201000%7D%7B100mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20acetic%20acid%7D%3D%5Cfrac%7B1%5Ctimes%20100%7D%7B1000%7D%3D0.1mol)
The chemical reaction for NaOH and acetic acid follows the equation:
![CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O](https://tex.z-dn.net/?f=CH_3COOH%2BNaOH%5Crightarrow%20CH_3COONa%2BH_2O)
<u>Initial:</u> 0.100 0.020
<u>Final:</u> 0.080 - 0.020
Volume of solution = 20 + 100 = 120 mL = 0.120 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of acetic acid = 4.74
![[CH_3COONa]=\frac{0.020}{0.120}](https://tex.z-dn.net/?f=%5BCH_3COONa%5D%3D%5Cfrac%7B0.020%7D%7B0.120%7D)
![[CH_3COOH]=\frac{0.080}{0.120}](https://tex.z-dn.net/?f=%5BCH_3COOH%5D%3D%5Cfrac%7B0.080%7D%7B0.120%7D)
pH = ?
Putting values in above equation, we get:
![pH=4.74+\log(\frac{0.020/0.120}{0.080/0.120})\\\\pH=4.14](https://tex.z-dn.net/?f=pH%3D4.74%2B%5Clog%28%5Cfrac%7B0.020%2F0.120%7D%7B0.080%2F0.120%7D%29%5C%5C%5C%5CpH%3D4.14)
Hence, the pH of the solution is 4.14