Chemistry 102: Acid Equilibrium: raising pH

Question

2. Consider the following reaction:

CH3​COOH(aq)+OH−(aq)CH3​COO−(aq)+H2​O(l)

How much 5.60 M NaOH must be added to 650.0 mL of a buffer that is 0.0185M acetic acid and 0.0265 M sodium acetate to raise the pH to 5.75?

Answer 1

It would help if we knew the Ka for CH3COOH. I will use 1.8x10^-5 as the value of Ka.

The pKa = -log Ka = -log 1.8x10^-5 = 4.74

Using the Henderson Hasselbalch equation

pH = pKa + log [CH3COO-] / [CH3COOH]

5.75 = 4.74 + log [CH3COO-] / [CH3COOH]

log [CH3COO-] / [CH3COOH] = 1.00

[CH3COO-] / [CH3COOH] = 10.1

CH3COOH + OH- ==> CH3COO-

0.0185 M......0.............0.0265..........Initial

-x.................-x.............+x.................Change

0.0185-x......0...............0.0265+x.....Equilibrium

0.0265+x / 0.0185-x = 10.1

x = 0.0144 M = [OH-]

(x ml)(5.60 M) = (600 ml)(0.0144 M)

x = 1.54 mls of 5.60 M NaOH must be added (this ignores the change in volume of 1.54 mls)

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Check:

0.0185 M CH3COOH - 0.0144 M = 0.0041 M CH3COOH final

0.0265 M CH3COO- + 0.0144 M = 0.0409 M CH3COO- final

pH = pKa + log [CH3COO-] / [CH3COOH]

pH = 4.74 + log (0.0409/0.0041) = 4.74 + log 9.98

pH = 4.74 + 0.999

pH = 5.74 Close enough to the desired 5.75