Question

Using the Henderson-Hasselbalch equation:

1. What is the calculated pH of a solution containing 5mL of 4.5% of Acetic acid (C₂H₄O₂) and 5mL of 0.500M of Sodium acetate(C₂H₃NaO₂)?

2. What is the calculated pH of a solution containing 5mL of 4.5% of Acetic acid (C₂H₄O₂) and 1mL of 0.500M of Sodium acetate (C₂H₃NaO₂)?

Assuming that the solutions of acetic acid and sodium acetate are equimolar.

HELP!!! Please show me how to do these calculations.

Answer 1

1. No of mole of CH3COONa    =   (5/1000)*(0.5) = 0.0025 mol

No of mol of CH3COOH   = (5*4.5/100)/60 = 0.00375 mol

pH = pka + log(salt/acid)

pka of CH3COOH = 4.74

pH = 4.74 + log(0.0025/0.00375) = 4.56

2.   No of mole of CH3COONa    =   (1/1000)*(0.5) = 0.0005 mol

No of mol of CH3COOH   = (5*4.5/100)/60 = 0.00375 mol

pH = pka + log(salt/acid)

pka of CH3COOH = 4.74

pH = 4.74 + log(0.0005/0.00375) = 3.865