A student is asked to determine the value of Ka for acetic acid by titration with potassium hydroxide. The student begins titrating a 46.4 mL sample of a 0.326 M aqueous solution of acetic acid with a 0.288 M aqueous potassium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 20.8 milliliters of potassium hydroxide have been added, the pH is 4.586. What is Ka for acetic acid based on the student's data?
No of mol of acetic acid taken = v*M = 46.4*0.326 = 15.13 mmol
No of mol of KOH added = 20.8*0.288 = 5.99 mmol
pH of acidic buffer = pka + log(base(or)salt/acid)
4.586 = x + log(5.99/(15.13-5.99))
x = pka = 4.77
ka = 10^-pka
= 10^-4.77
= 1.7*10^-5
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