Respuesta :
Answer: The pH of resulting solution is 9.08
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] ........(1)
- For HCl:
Molarity of HCl = 0.40 M
Volume of solution = 15.0 mL
Putting values in equation 1, we get:
[tex]0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol[/tex]
- For ammonia:
Molarity of ammonia = 0.50 M
Volume of solution = 20.0 mL
Putting values in equation 1, we get:
[tex]0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol[/tex]
The chemical reaction for hydrochloric acid and ammonia follows the equation:
[tex]HCl+NH_3\rightarrow NH_4Cl[/tex]
Initial: 0.006 0.01
Final: - 0.004 0.006
Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L (Conversion factor: 1 L = 1000 mL)
- To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[salt]}{[base]})[/tex]
[tex]pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})[/tex]
We are given:
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of ammonia = [tex]-\log (1.8\times 10^{-5})=4.74[/tex]
[tex][NH_4Cl]=\frac{0.006}{0.035}[/tex]
[tex][NH_3]=\frac{0.004}{0.035}[/tex]
pOH = ?
Putting values in above equation, we get:
[tex]pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-4.92=9.08[/tex]
Hence, the pH of the solution is 9.08
