Question
In: Chemistry
Given 100.0 ml of a buffer that is 0.50 M in HOCl and 0.40 M in...
Given 100.0 ml of a buffer that is 0.50 M in HOCl and 0.40 M in NaOCl, what is the pH after 10.0 ml of 1.0 M NaOH has been added?
(Ka for HOCl =3.5 x 10-8)
Solutions
Expert Solution
Pka = -logka
= -log3.5*10-8
= 7.4559
no of moles of HOCl = molarity * volume in L
= 0.5*0.1 = 0.05 moles
no of moles of NaOCl = 0.4*0.1 = 0.04 moles
PH = PKa + log[NaOCl]/[HOCl]
= 7.4559 + log0.04/0.05
= 7.4559-0.0969 = 7.359
By the addition of NaOH
no of moles of NaOH = 1*0.01 = 0.01 moles
no of moles of HOCl = 0.05-0.01 = 0.04 moles
no of moles of NaOCl = 0.04+0.01 = 0.05 moles
PH = Pka + log[NaOCl]/[HOCl]
= 7.4559 + log0.05/0.04
= 7.4559 + 0.09691 = 7.5528
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