Final answer:
To find the pH of a solution prepared by mixing Sodium Acetate and HCl, a neutralization reaction and the Henderson-Hasselbalch equation are used. After calculations, the pH is approximately 5.35.
Step-by-step explanation:
To calculate the pH of a solution made by mixing 500 ml of 0.25 M Sodium Acetate with 250 ml 0.1 M HCl, we need to consider the acid-base neutralization reaction that occurs between the acetic acid portion of the sodium acetate and the HCl. A buffer solution will form as acetate ions from sodium acetate react with the hydrogen ions from HCl to produce acetic acid. In this reaction, since sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide), and HCl is a strong acid, it will donate protons to the acetate ions, reducing the pH of the solution slightly but maintaining a buffer capacity.
First, we find the moles of HCl and sodium acetate. Since moles = concentration × volume, for HCl it is: 0.1 M × 0.250 L = 0.025 moles and for Sodium Acetate it is: 0.25 M × 0.5 L = 0.125 moles. After the reaction, there will be excess sodium acetate since HCl is the limiting reagent. The moles of remaining sodium acetate, i.e., acetate ions will be 0.125 moles - 0.025 moles = 0.1 moles.
Then, we use the Henderson-Hasselbalch equation to find the pH: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of acetic acid and [A-] and [HA] are the concentrations of acetate and acetic acid respectively. To find the concentrations, we divide the moles by the total volume of the solution in liters.
Assuming the pKa of acetic acid is 4.75, the pH calculation would be as follows (note that the [HA] would be equivalent to the moles of HCl used, 0.025 moles, and the total volume 0.75 L):
pH = 4.75 + log ([0.1]/[0.025])
pH = 4.75 + log (4)
pH ≈ 4.75 + 0.602
pH ≈ 5.35
Therefore, the pH of the solution after mixing is approximately 5.35.