Question

1. In Activity 2, there was 2 mL of buffer per well. Calculate the final pH of the undiluted buffer after 0.25 mL of 0.1 M NaOH are added. The Ka of CH3COOH = 1.8 × 10-5. Use the CH3COOH buffer concentration calculated in Data Table 2(1.1 M), a balanced CH3COOH + NaOH reaction equation, an ICE table, and a balanced dissociation equilibrium equation to help determine your pH. 2. What is the final pH of the second dilution after adding 1 drop of 0.1 M NaOH?

Answer 1

Initial concentration of acid[CH3COOH] and base [CH3COO-] in the buffer solution

Using Hendersen-Hasselbalck equation,

Since the initial pH is not given, we would assume it to be 3.0

pH = pKa + log(base/acid)

3.0 = 4.75 + log([CH3COO-]/[CH3COOH])

[CH3COO-] = 0.02[CH3COOH]

[CH3COOH] + [CH3COO-] = 1.1 M

[CH3COOH] + 0.02[CH3COOH] = 1.1

[CH3COOH] = 1.1/1.02 = 1.08 M

[CH3COO-] = 1.1 - 1.08 = 0.02 M

After NaOH is added,

CH3COOH + NaOH = CH3COONa + H2O

NaOH = 0.1 M x 0.25 ml = 0.025 mmol

final [CH3COOH] = 1.08 - 0.025 = 1.055 mmol

final [CH3COO-] = 0.02 + 0.025 = 0.045 mmol

ICE chart

                CH3COOH + NaOH ---> CH3COONa + H2O

I                   1.08           0.025                -                   

C               -0.025          -0.025              0.025

E                1.055              0                  0.025

So,

pH = 4.75 + log(0.045/1.055) = 3.38

[pl. note, correct value of initial pH is required to get the right answer, feed the value and repeat the calculation as shown above]