Question

NaOH is added to a 500 mL of 2M acetic acid. If the pKa value of acetic acid is approximately 4.8, what volume of 2M NaOH must be added so that the pH of the solution is 4.8?

Answer 1

To achieve a pH of 4.8, approximately 166.67 mL of 2M NaOH must be added to the 500 mL of 2M acetic acid solution.

Step-by-step explanation:

To calculate the volume of 2M NaOH that must be added to the 500 mL of 2M acetic acid solution to achieve a pH of 4.8, we can use the Henderson-Hasselbalch equation:

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

Where [A-] and [HA] represent the concentrations of the acetate ion and acetic acid, respectively. In this case, we want pH = 4.8 and pKa = 4.8. Rearranging the equation, we get:

[A-]/[HA] = 10^(pH - pKa)

Since we used 500 mL of 2M acetic acid, we have 1 mol of acetic acid. So, the initial concentration of acetic acid is 1M. Let's represent the volume of 2M NaOH to be added as V mL.

So, we have:

2M * V = 1M * (500 mL + V)

Simplifying, we find:

V = 500/3 = 166.67 mL

Therefore, approximately 166.67 mL of 2M NaOH must be added to achieve a pH of 4.8.