Respuesta :
The question is incomplete, here is the complete question:
An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid with a solution of 0.8400 M KOH. The [tex]pK_a[/tex] of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added. Round your answer to 2 decimal places.
Answer: The pH of the solution is 2.69
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 nitrous acid:
Molarity of nitrous acid = 1.200 M
Volume of solution = 182.2 mL
Putting values in above equation, we get:
[tex]1.200M=\frac{\text{Moles of nitrous acid}\times 1000}{182.2mL}\\\\\text{Moles of nitrous acid}=\frac{1.200\times 182.2}{1000}=0.219mol[/tex]
- For KOH:
Molarity of KOH = 0.8400 M
Volume of solution = 46.44 mL
Putting values in above equation, we get:
[tex]0.8400M=\frac{\text{Moles of KOH}\times 1000}{46.44mL}\\\\\text{Moles of KOH}=\frac{0.8400\times 46.44}{1000}=0.039mol[/tex]
The chemical reaction for KOH and nitrous acid follows the equation:
[tex]HNO_2+KOH\rightarrow KNO_2+H_2O[/tex]
Initial: 0.219 0.039
Final: 0.18 - 0.039
Volume of solution = 182.2 + 46.44 = 228.64 mL = 0.22864 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{[KNO_2]}{[HNO_2]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of nitrous acid = 3.35
[tex][KNO_2]=\frac{0.039}{0.22864}[/tex]
[tex][HNO_2]=\frac{0.18}{0.22864}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=3.35+\log(\frac{0.039/0.22864}{0.18/0.22864})\\\\pH=2.69[/tex]
Hence, the pH of the solution is 2.69
