Question

# Calculate the pH at the equivalence point in titrating 0.093 M solutions of each of the...

Calculate the pH at the equivalence point in titrating 0.093 M solutions of each of the following with 0.098 M NaOH.

(a) hydrobromic acid (HBr)
pH =

(b) phenol (HC6H5O), Ka = 1.3e-10
pH =

(c) ascorbic acid (HC6H7O6), Ka = 8e-05

pH =

(a) HBr is a strong acid . A equivalence point pH will be 7

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at equivalence point all the phenol is converted to C6H5ONa. It will hydrolyse back to phenol.

C6H5O^- + H2O <==> C6H5OH + OH-

 C6H5O^- C6H5OH H+ initial 0.093 0 0 change -x +x +x equilibrium 0.093-x +x +x

Kb = 10^-14/Ka = 10^-14/1.3*10^-10 = 0.77*10^-4

Kb = [OH-][C6H5OH]/[C6H5O-]

or, 0.77*10^-4 = x^2/0.093-x

or, 0.072 *10^-4 -0.77*10^-4x -x ^2 = 0

or, x = 2.64*10^-3

[OH-] = 2.64*10^-3

pOH = -log [H+] = -log [2.64*10^-3] = 2.58

pH = 14-pOH = 11.42

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Kb of ascorbate = Kw/Ka = 1*10^-14/8*10^-5 =0.125 *10^-9

Kb = [OH-][ascorbate]/[ascorbic acid]

0.125 *10^-9 = x^2/0.093-x

or, 0.012*10^-9 -0.125*10^-9x -x^2 = 0

or, x = 3.46*10^-6

[OH-] = 3.46*10^-6

pOH = -log[OH-] = 5.46

pH = 14-pOH = 14- 5.46 = 8.54

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