Question

100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M CH₃COOH solution. The pH of the resulting solution will be (pkₐ = 4.74)

Answer 1

The pH of the resulting solution will be 4.74.

Step-by-step explanation:

The pH of the resulting solution can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Here, CH₃COOH is the weak acid (HA) and NaOH is the strong base that reacts with it, forming CH₃COO- and H₂O. Since the initial concentration of CH₃COOH is 0.2 M and its pKa is 4.74, we can substitute these values into the equation:

pH = 4.74 + log([CH₃COO-]/[CH₃COOH])

Since we are adding 0.1 M NaOH, the concentration of CH₃COO- will increase by 0.1 M, while the concentration of CH₃COOH will decrease by 0.1 M. Substituting these values into the equation:

pH = 4.74 + log(0.1/0.1) = 4.74 + log(1) = 4.74 + 0 = 4.74

Therefore, the pH of the resulting solution will be 4.74.