Question

1) What is the pH of 100.0 mL of a 0.100M aqueous solution of benzoic acid at 25C?

2) What is the pH of 100.0mL of a 0.100M aqueous solution of sodium benzoate at 25C?

3) What is the pH of a mixture of 50.0mL of 0.100M benzoic acid and 50.0mL of 0.100M sodium benzoate at 25C?

Answer 1

1) pH of weak acid(benzoicacid) = 1/2(pka-logC)

           pka of benzoicacid = 4.2

    pH = 1/2(4.2-log0.1) = 2.6

2) pH of sodium benzoate = 7 +1/2(pka+logC)

    pka of benzoicacid = 4.2

C = concentration of sodium benzoate = 0.1 M

pH = 7+1/2(4.2+log0.1) = 8.6

3) no of mol of benzoicacid = 50*0.1 = 5 mmol

no of mol of sodium benzoate =   50*0.1 = 5 mmol

pH = pka + log(sodium benzoate/benzoicacid)

   as both components , no of moles same.

pH = pka = 4.2