Respuesta :
Answer: The pH of resulting solution is 3.815
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 lactic acid:
Molarity of lactic acid solution = 0.100 M
Volume of solution = 25.0 mL
Putting values in above equation, we get:
[tex]0.100M=\frac{\text{Moles of lactic acid}\times 1000}{25mL}\\\\\text{Moles of lactic acid}=\frac{0.100\times 25}{1000}=0.0025mol[/tex]
- For NaOH:
Molarity of NaOH solution = 0.050 M
Volume of solution = 23.5 mL
Putting values in above equation, we get:
[tex]0.050M=\frac{\text{Moles of NaOH}\times 1000}{23.5mL}\\\\\text{Moles of NaOH}=\frac{0.050\times 23.5}{1000}=0.0012mol[/tex]
The chemical reaction for NaOH and lactic acid follows the equation:
[tex]HC_3H_5O_3+NaOH\rightarrow C_3H_5O_3Na+H_2O[/tex]
Initial: 0.0025 0.0012
Final: 0.0013 - 0.0012
Volume of solution = 25 + 23.5 = 48.5 mL = 0.0485 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_3H_5O_3Na]}{[HC_3H_5O_3]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of lactic acid = 3.85
[tex][HC_3H_5O_3]=\frac{0.0013}{0.0485}[/tex]
[tex][C_3H_5O_3Na]=\frac{0.0012}{0.0485}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=3.85+\log(\frac{0.0012/0.0485}{0.0013/0.0485})\\\\pH=3.815[/tex]
Hence, the pH of resulting solution is 3.815
