Question

Consider the titration of 120.0 mL of 0.018 M HOCl (Ka=3.5×10−8) with 0.0440 M NaOH.
a.How many milliliters of 0.0440 M NaOH are required to reach the equivalence point?
b.Calculate the pH after the addition of 12.0 mL of 0.0440 M NaOH.
c.Calculate the pH halfway to the equivalence point.
d.Calculate the pH at the equivalence point.

Answer 1

Ka = 3.5*10^-8

pKa = - log(Ka)
pKa = - log(3.5*10^-8)
pKa = 7.46


(a) volume of NaOH to reach the equivalence point ?
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Ca x Va = Cb x Vb

Vb = Ca x Va / Cb
Vb = 0. 018 x 120 / 0.0440
Vb = 49.0909 mL of NaOH solution . . . !



(d) pH at the equivalence point ?
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Csalt = Ca x Va / (Va + Vb)
Csalt = 0.018 x 120 / (120 + 49.09)
Csalt = 0.01277 mol/L


pH = 1/2 (pKw + pKa + log(Csalt))
pH = 1/2 (14 + 7.46 + log(0.01277)
pH = 9.78


(c) pH of the point halfway between the beginning of the titration and the equivalence point ?
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pH(beginning) = pKa / 2 - log(Ca) / 2
pH(beginning) = 7.46 / 2 - log(0.018) / 2
pH(beginning) = 4.60

At halfway : pH = pKa + log( [ClO(-)] / [HClO] )

and [ClO(-)] = [HClO]

. . . therefore :

pH(halfway) = pKa = 7.46


(b) pH after addition of 12 mL titration solution ?
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[ClO(-)] = ( Ca x Va + Cb x Vb ) / (Va + Vb)
[ClO(-)] = ( 0.018 x 120 + 0.0440 x 12) / (120+12)
[ClO(-)] = 0. 02036 mol/L


[HClO] = ( Ca x Va - Cb x Vb ) / (Va + Vb)
[HClO] = ( 0.018 x 120 - 0.0440 x 12) / 132
[HClO] = 0.01236 mol/L


pH = pKa + log( [ClO(-)] / [HClO] )
pH = 7.46 + log(0.02036 / 0.01236)
pH = 7.67