N-Acetyltryptophan (Ka=4.17 x 10^-3) was titrated with Sodium hydroxide. In the

Question

N-Acetyltryptophan (Ka=4.17 x 10^-3) was titrated with Sodium hydroxide. In the titration, 125.0 mL of 1.00 x 10^-2 M N-acetyltryptophan was titrated with 0.0250 M NaOH. The temperature for the entire titration was maintained at 25^o C. Answer questions A-E:

A. What is the pH of the original N-acetyltryptophan solution prior to the addition of NaOH?

B. What is the pH of this solution if 30.0 mL of the NaOH was added?

C. At which pH will the equivalence point be observed?

D. What is the pH after 75.0 mL of the NaOH has been added?

E. At which pH will the midpoint be observed?

Explanation / Answer

pka = -logka = -log(4.17*10^-3) = 2.38

A. pH = 1/2(pka-logC)

      = 1/2(2.38-log0.01) = 2.19

B. No of mol of N-acetyltryptophan = 0.125*0.01 = 0.00125 mol

   No of mol of NaOH = 0.03*0.025 = 0.00075 mol

pH = pka + log(NaOH/acid)

     = 2.38+log(0.00075/(0.00125-0.00075))

     =2.56

C. At equivalence point ,

No of mol of acid = No of mol of NaOH

volume of NaOH required = 0.00125/0.025 = 0.05 L = 50 ml

concentration of salt = 0.00125/(0.175) = 0.0071 M

pH = 7+1/2(pka+log C)

   = 7+1/2(2.38+log0.0071)

    = 7.11

d. concentration of excess NaOH = (75-50)/200*0.025 = 0.003125 M

pH = 14-(-log(0.003125))

     = 11.5

E. at midpoint pH = pka = 2.38

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