Respuesta :
Answer: The new pH of resulting solution is 6.03
Explanation:
We are adding hydrochloric acid to the solution, so it will react with salt (sodium hydrogen carbonate) only.
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 hydrochloric acid:
Molarity of hydrochloric acid = 0.15 M
Volume of solution = 3 mL
Putting values in above equation, we get:
[tex]0.15M=\frac{\text{Moles of hydrochloric acid}\times 1000}{3mL}\\\\\text{Moles of hydrochloric acid}=0.00045mol[/tex]
- For sodium hydrogen carbonate:
Molarity of sodium hydrogen carbonate = 0.025 M
Volume of solution = 150 mL
Putting values in above equation, we get:
[tex]0.025M=\frac{\text{Moles of sodium hydrogen carbonate}\times 1000}{150mL}\\\\\text{Moles of sodium hydrogen carbonate}=0.00375mol[/tex]
- For carbonic acid:
Molarity of carbonic acid = 0.045 M
Volume of solution = 150 mL
Putting values in above equation, we get:
[tex]0.045M=\frac{\text{Moles of carbonic acid}\times 1000}{150mL}\\\\\text{Moles of carbonic acid}=0.00675mol[/tex]
The chemical reaction for sodium hydrogen carbonate and hydrochloric acid follows the equation:
[tex]NaHCO_3+HCl\rightarrow NaCl+H_2CO_3[/tex]
Initial: 0.00375 0.00045 0.00675
Final: 0.00330 - 0.00720
Volume of solution = 150 + 3 = 153 mL = 0.153 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{[NaHCO_3]}{[H_2CO_3]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of carbonic acid = 6.37
[tex][NaHCO_3]=\frac{0.0033}{0.153}[/tex]
[tex][H_2CO_3]=\frac{0.0072}{0.153}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=6.37+\log(\frac{0.0033/0.153}{0.0072/0.153})\\\\pH=6.03[/tex]
Hence, the new pH of the solution is 6.03
