Answer:
pH = 5.01
Explanation:
The reaction between the sodium propanoate and the HCl added is the following:
CH₃CH₂COO⁻ + H₃O⁺ ⇄ CH₃CH₂COOH + H₂O
initial 0.110M 0.061moles 0.253M
The number of moles of the acid propanoic (a) and sodium propanoate (b) is:
[tex]\eta_{a} = [a]*Va = 0.110 M * 1.41 L = 0.155 moles \thinspace acid[/tex]
[tex]\eta_{b} = [b]*Vb = 0.253 M * 1.41 L = 0.357 moles\thinspace propanoate[/tex]
After the adding of HCl, the number of moles of acid propanoic and propanoate is:
[tex]\eta_{b} = 0.357 moles - 0.061 moles = 0.296 moles \thinspace propanoate[/tex]
[tex]\eta_{a} = 0.155 moles + 0.061 moles = 0.216 moles \thinspace acid[/tex]
Hence, the pH of the solution after the addition of HCl is:
[tex] pH = pKa + log(\frac{b}{a}) = -log(1.34 \cdot 10^{-5}) + log(\frac{0.296}{0.216}) = 5.01 [/tex]
Therefore, the pH of the solution is 5.01.
I hope it helps you!
