eneral Chemistry 4th Edition University Science Books presented by Sapling Learn
ID: 1031047 • Letter: E
Question
eneral Chemistry 4th Edition University Science Books presented by Sapling Learning Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.180 M HCIO (aq) with 0.180 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 40.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH Previous & Give Up & View SolutionCheck Answer Next HintExplanation / Answer
a) before addition of KOH
pka of HOCl = 7.53
pH of weak acid = 1/2(pka-logC)
C = concentration of acid = 0.18 M
pH = 1/2(7.53-log0.18)
= 4.14
b) pka of HOCl = 7.53
no of mol of HOCl = 50*0.18 = 9 mmol
no of mol of koh = 25.0*0.18 = 4.5 mmol
pH = pka + log(KOCl/HOCl)
= 7.53 + log(4.5/(9-4.5))
= 7.53
C) pka of HOCl = 7.53
no of mol of HOCl = 50*0.18 = 9 mmol
no of mol of koh = 40*0.18 = 7.2 mmol
pH = pka + log(KOCl/HOCl)
= 7.53 + log(7.2/(9-7.2))
= 8.13
d)
no of mol of HOCl = 50*0.18 = 9 mmol
no of mol of koh = 50*0.18 = 9 mmol
total HOCl Reacts with KOH and gives KOCl
concentration of KOCl = n/v = 9/100 = 0.09 M
pH of KOCl = 7+1/2(pka+logC)
pka of HOCl = 7.53
concentration of KOCl = 0.09 M
= 7+1/2(7.53+log0.09)
= 10.24
e) no of mol of HOCl = 50*0.18 = 9 mmol
no of mol of koh = 60*0.18 = 10.8 mmol
excess KOH = 10.8-9 = 1.8 mmol
concentration of excess KOH = 1.8/110 = 0.016 M
pOH = -log(OH-)
= -log(0.016)
= 1.8
pH = 14-1.8 = 12.2
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