Respuesta :
Answer: The pH of the solution is 4.14
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
- For NaOH:
Molarity of NaOH = 1 M
Volume of solution = 20 mL
Putting values in above equation, we get:
[tex]1M=\frac{\text{Moles of NaOH}\times 1000}{20mL}\\\\\text{Moles of NaOH}=\frac{1\times 20}{1000}=0.02mol[/tex]
- For acetic acid:
Molarity of acetic acid solution = 0.26 M
Volume of solution = 100 mL
Putting values in above equation, we get:
[tex]1M=\frac{\text{Moles of acetic acid}\times 1000}{100mL}\\\\\text{Moles of acetic acid}=\frac{1\times 100}{1000}=0.1mol[/tex]
The chemical reaction for NaOH and acetic acid follows the equation:
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]
Initial: 0.100 0.020
Final: 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:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.74
[tex][CH_3COONa]=\frac{0.020}{0.120}[/tex]
[tex][CH_3COOH]=\frac{0.080}{0.120}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=4.74+\log(\frac{0.020/0.120}{0.080/0.120})\\\\pH=4.14[/tex]
Hence, the pH of the solution is 4.14
