A 1.41 L buffer solution consists of 0.253 M propanoic acid and 0.110 M sodium propanoate. Calculate the pH of the solution following the addition of 0.061 mol HCl . Assume that any contribution of the HCl to the volume of the solution is negligible. The K a of propanoic acid is 1.34 × 10 − 5 .
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Ответ:
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:
After the adding of HCl, the number of moles of acid propanoic and propanoate is:
Hence, the pH of the solution after the addition of HCl is:
Therefore, the pH of the solution is 5.01.
I hope it helps you!
Ответ:
m₀=60 g
w=100m₁/m₀
w=100*50/60=83.3%