Question

Calculate the pH at the equivalence point in titrating 0.034 M solutions of each of the following with 0.093 M NaOH. (a) periodic acid (HIO4) pH = (b) hydrocyanic acid (HCN), Ka = 4.9e-10 pH = (c) arsenous acid (H3AsO3), Ka = 5.1e-10

Answer 1

a) 0.034 M HIO4 with 0.093 M NaOH

  HIO4 is a strong acid and NaOH is a strong base.

Therefore, at equivalence point, pH =7.

b) 0.034 M HCN ( Ka= 4.9 x 10^-10) with 0.093 M NaOH

      NaOH        +             HCN           ------------->     NaCN     +    H2O

   0.093 M                       0.034 M                              0

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0.093 - 0.034                   0                                    0.034 M

= 0.059 M

Since NaCN formed from strong base NaOH and weak acid HCN, NaCN is basic.

      [NaOH] = 0.059 M

      [HCN] = 0.034 M

Given that Ka of HCN = 4.9 x 10^-10

Kw = Ka.Kb

Kb = Kw/ Ka = 1.0 x 10^-14 / 4.9x 10^-10 = 0.204 x 10^-4

pOH = - logKb+log [NaCN]/ [NaOH]

    = - log (0.204 x 10^-4) + log (0.034 M/0.059 M)

   = 4.45

pOH = 4.45

pH = 14 - pOH = 14-4.45 = 9.55

pH = 9.55

c) 0.034 M H3AsO3 ( Ka= 5.1 x 10^-10) with 0.093 M NaOH

      3NaOH        +             H3AsO3           ------------->     Na3AsO3     +    3H2O

    3x 0.093 M                     0.034 M                                     0

= 0.279 M

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0.279 - 0.034                   0                                           0.034 M

= 0.245 M

Since Na3AsO3 formed from strong base NaOH and weak acid H3AsO3 , Na3AsO3 is basic.

      [NaOH] = 0.245 M

      [Na3AsO3] = 0.034 M

Given that Ka of Na3AsO3 = 5.1x 10^-10

Kw = Ka.Kb

Kb = Kw/ Ka = 1.0 x 10^-14 / 5.1x 10^-10 = 0.196 x 10^-4

pOH = - logKb+log [Na3AsO3]/ [NaOH]

    = - log (0.196 x 10^-4) + log (0.034 M/0.245 M)

   = 3.85

pOH = 3.85

pH = 14 - pOH = 14-3.85 = 10.15

pH = 10.15