Determine the pH during the titration of 66.6 mL of 0.468 M hypochlorous acid (Ka =...

Determine the pH during the titration of 66.6 mL of 0.468 M hypochlorous acid (K_a = 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 ?

Solutions

Expert Solution

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

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

queen_honey_blossom answered 1 year ago

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