A solution of ethanoic acid in water has a concentration of 3 g/dm^3. Given that the pKa of ethanoic acid is 4.76, calculate the pH of this solution.
Key facts to be noted by student:
Molar mass of ethanoic acid = 60 g/mol.
pKa = -log(Ka)
pH =-log([H+])
Concentration of acid solution = 3/60 = 0.05 moldm-3
ka = [H+]2 / [CH3COOH] therefore
[H+]2 = 10-4.76 / 0.05 --> [H+] = 0.0186 moldm-3
ph = -log(0.0186) ---> pH =1.73
TC
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