Respuesta :
Answer: The pH of the solution is 4.57
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 benzoic acid:
Molarity of benzoic acid = 50 mM = 0.05 M (Conversion factor: 1 M = 1000 mM)
Volume of solution = 100 mL
Putting values in above equation, we get:
[tex]0.05M=\frac{\text{Moles of benzoic acid}\times 1000}{100mL}\\\\\text{Moles of benozic acid}=\frac{(0.05\times 100)}{1000}=0.005mol[/tex]
- For sodium hydroxide:
Molarity of sodium hydroxide = 50 mM = 0.05 M
Volume of solution = 70 mL
Putting values in above equation, we get:
[tex]0.05M=\frac{\text{Moles of sodium hydroxide}\times 1000}{50mL}\\\\\text{Moles of sodium hydroxide}=\frac{0.05\times 70}{1000}=0.0035mol[/tex]
The chemical reaction for sodium hydroxide and benzoic acid follows the equation:
[tex]C_6H_5COOH+NaOH\rightarrow C_6H_5COONa+H_2O[/tex]
Initial: 0.005 0.0035
Final: 0.0015 - 0.0035
Volume of solution = 100 + 70 = 170 mL = 0.170 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{[C_6H_5COONa]}{[C_6H_5COOH]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of benzoic acid = 4.2
[tex][C_6H_5COONa]=\frac{0.0035}{0.170}[/tex]
[tex][C_6H_5COOH]=\frac{0.0015}{0.170}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=4.2+\log(\frac{0.0035/0.170}{0.0015/0.170})\\\\pH=4.57[/tex]
Hence, the pH of the solution is 4.57
