What is the pH of a solution containing 0.10 M propanoic acid (HC3H5O2) and 0.10 M sodium propanoate after 0.02 mol of HCl is added to 1.0 L of the solution?
Use the Henderson-Hasselbalch equation
To find the pH of the solution, we need to understand the concept of acid-base equilibrium and the dissociation of propanoic acid.
Propanoic acid (HC3H5O2) is a weak acid and can dissociate in water according to the following equation:
HC3H5O2 ⇌ H+ + C3H5O2-
Sodium propanoate (C3H5O2Na) is the conjugate base of propanoic acid, and it can dissociate to release sodium ions and the conjugate base:
C3H5O2Na → Na+ + C3H5O2-
When HCl is added to the solution, it will react with the propanoate ions to form more propanoic acid:
HCl + C3H5O2- → HC3H5O2 + Cl-
Since we know the initial concentrations of propanoic acid and sodium propanoate are both 0.10 M, the amount of propanoic acid (HC3H5O2) will increase while the amount of propanoate ions (C3H5O2-) will decrease.
Now, let's calculate the new concentrations of propanoic acid and propanoate ions after the addition of HCl:
The amount of propanoic acid (HC3H5O2) initially is 0.10 M × 1.0 L = 0.10 mol.
The amount of propanoate ions (C3H5O2-) initially is 0.10 M × 1.0 L = 0.10 mol.
The amount of HCl added is 0.02 mol.
Since HCl reacts with propanoate ions in a 1:1 ratio, the amount of propanoate ions that reacted with HCl is also 0.02 mol.
Now, let's calculate the final concentrations of propanoic acid and propanoate ions:
The final amount of propanoic acid (HC3H5O2) is 0.10 mol + 0.02 mol = 0.12 mol.
The final amount of propanoate ions (C3H5O2-) is 0.10 mol - 0.02 mol = 0.08 mol.
To calculate the pH of the solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([conjugate base] / [acid])
The pKa value for propanoic acid is typically given as 4.87.
Using the equation, we can find:
pH = 4.87 + log(0.08 mol / 0.12 mol)
Calculating the ratio inside the logarithm:
pH = 4.87 + log(0.66)
Calculating the logarithm:
pH = 4.87 + (-0.18)
Finally:
pH ≈ 4.69
The pH of the solution containing 0.10 M propanoic acid and 0.10 M sodium propanoate after 0.02 mol of HCl is added is approximately 4.69.