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ID: 556837 • Letter: I
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ibis/view.php?id-3858971 Jump to. 11/8/2017 11:30 PM 0 13.9/15 Question 9 of 14 General ChemistryEdn niversity Science Books Calculace the ph for each of the folowing cases in the titration of 50.0 mL of 0.160 M HCIO) with 0.160 M KOH(aq) The ionization constant for HCIO can be found here. (a) belore addition of any KOH (b) ater addition of 25.0 m of KOH (c) afer addision of3.0of KOH D (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH about us careers privacy policy tes of use contact us 0 4Explanation / Answer
pka of HClO = -log(3*10^-8) = 7.52
a) befor addition of koh
pH = 1/2(pka-logC)
= 1/2(7.52-log0.16)
= 4.16
b) after addition of 25 ml KOH
no of mol of HClO = 50*0.16 = 8 mmol
no of mol of KOH = 25*0.16 = 4 mmol
at half equivalence point, pH = pka = 7.52
c) after addition of 35 ml KOH
no of mol of HClO = 50*0.16 = 8 mmol
no of mol of KOH = 35*0.16 = 5.6 mmol
pH = pka + log(base/acid-base)
= 7.52+log(5.6/(8-5.6))
= 7.9
d) after addition of 50 ml KOH
no of mol of HClO = 50*0.16 = 8 mmol
no of mol of KOH = 50*0.16 = 8 mmol
at equivalence point,
concentration of salt = 8/100 = 0.08 M
pH = 7+1/2(pka+logC)
= 7+1/2(7.52+log0.08)
= 10.21
e) concentration of excess KOH = (60-50)*0.16/110 = 0.0145 M
pH = 14 - (-log(OH-))
= 14 - (-log0.0145)
= 12.16
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