In: Chemistry
Determine the pH during the titration of 66.6
mL of 0.468 M hypochlorous acid
(Ka = 3.5×10-8) by
0.468 M NaOH at the following
points. (Assume the titration is done at 25 °C.)
(a) Before the addition of any NaOH ?
(b) After the addition of 16.0 mL of
NaOH ?
(c) At the half-equivalence point (the titration midpoint) ?
(d) At the equivalence point ?
(e) After the addition of 99.9 mL of
NaOH ?
pKa = -logKa = = -log(3.5 x10^ -8.) = 7.46
millimoles of HClO= 66.6 x 0.468 = 31.2
a) before addition of NaOH
pH = 1/2 (pKa- log C)
= 1/2 (7.46 -log (0.468) ) = 3.90
pH= 3.90
(b) After the addition of 16.0 mL of NaOH
millimoles of NaOH = 16 x 0.468 = 7.49
HOCl + NaOH -------------------> NaOCl + H2O
31.2 7.49 0 0
23.71 0 7.49
pH = pKa + log [NaOCl / HOCl]
pH = 6.96
(c) at half equivalence
here pH = pKa
pH = 7.46
d ) at equilvalence point :
here only salt remains = salt concentration = 31.2 / (66.6 + 66.6) = 0.234 M
pH = 7 + 1/2 [pKa + logC]
pH = 7 + 1/2 [7.46 + log0.234]
pH = 10.42
(e) After the addition of 99.9 mL of NaOH ?
pH = 12.97