Question

50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) is titrated with a 0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d. After the addition of 50.00mL of NaOH e. After the addition of 51.00mL of NaOH. PLEASE EXPLAIN STEPS AND NOT JUST SHOW WORK, I KNOW THE FIRST ONE, THE OTHERS ARE GIVING ME TROUBLE

Answer 1

a. Before Titration

   pH = 1/2(pka-logC)

      = 1/2(4.72-log0.1)

      = 2.86

b. After the addition of 5.00mL of NaOH

no of mole of HN3 = 50*0.1/1000 = 0.005 mole

no of mole of NaOH = 5*0.1/1000 = 0.0005 mole

   pka of HN3 = 4.72

pH = pka + log(base/acid)

      = 4.72 +log(0.0005/(0.005-0.0005))

      = 3.76

c. After the addition of 25.00mL of NaOH

    no of mole of HN3 = 50*0.1/1000 = 0.005 mole

no of mole of NaOH = 25*0.1/1000 = 0.0025 mole

   pka of HN3 = 4.72

pH = pka + log(base/acid)

      = 4.72 +log(0.0025/(0.005-0.0025))

       = 4.72

d. After the addition of 50.00mL of NaOH

    no of mole of HN3 = 50*0.1/1000 = 0.005 mole

no of mole of NaOH = 50*0.1/1000 = 0.005 mole

concentration of salt formed = 0.005/100*1000 = 0.05 M

reaches to equivalencepoint

pH = 7+1/2(pka+logC)

     = 7+1/2(4.72+log0.05)

     = 8.71

e. After the addition of 51.00mL of NaOH

   no of mole of NaOH excess = (51-50)*0.1/1000 = 0.0001 mole

concentration of NaOH excess = 0.0001/101*1000 = 0.00099 M

   pH = 14 - (-log(OH-))

      = 14 - (-log(0.00099))

      = 11