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+]/ [CH3COOH]   therefore

[H+]2 = 10-4.76 / 0.05   --> [H+] = 0.0186 moldm-3

ph = -log(0.0186)   ---> pH =1.73

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