A reminder of the Henderson-Hasselbalch equation: pH = pK_a + log [salt form]/[a

Question

A reminder of the Henderson-Hasselbalch equation: pH = pK_a + log [salt form]/[acid form] Do the calculations to make 200mL of 0.1M sodium acetate buffer with a pH of 5.1, starting with 5.0M acetic acid and 1.0M NaOH. Calculate the total amount of acetic acid needed. Calculate the ratio of the two forms of acetate (A^- and HA) that will exist when the pH is 5.1. Use this ratio to calculate the % of acetate that will be in the A^- form. Assume that each NaOH will convert one HAc to an Ac. Use this plus the % A^- to calculate the amount of NaOH needed to convert the correct amount of HAc to Ac^-. How much acetic acid is needed What is the ratio of Ac- to HAc at pH of 5.1 What fraction of total acetate is Ac- at pH 5.1 How much OH- is needed to obtain 68.7% Ac-

Explanation / Answer

pH = pKa + log(base/acid)

pKa = 4.74

pH = 5.1

so,

5.1 = 4.74 + log([CH3COO-]/[CH3COOH])

[CH3COO-] = 2.29[CH3COOH]

[CH3COOH] + [CH3COO-] = 0.1 M x 200 ml = 20 mmol

[CH3COOH] + 2.29[CH3COOH] = 20

[CH3COOH] = 6.08 mmol

1. Amount of acetic acid needed = 6.08 mmol/5 M = 1.216 ml of 5 M

2. ratio of [Ac-]/[HAc] at pH 5.1 = 2.29

3. fraction of Ac- at pH 5.1 = ((20-6.08)/20) x 100 = 69.6%

4. for 68.7% Ac-,

68.7 % of 20 mmol = 13.74 mmol

[OH-] required = 13.74 mmol/1 M = 13.74 ml of 1.0 M NaOH

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