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Grade: 12th pass
        
#1problems:
a titration is performed by adding 0.600M KOH to 40.0 ML of 0.800 M HCL..
  1. calculate the ph before addition of any koh..
  2. calculate the ph after addition of 5.0 ML of the base..
5 years ago

Answers : (1)

Sumit Majumdar
IIT Delhi
askIITians Faculty
137 Points
							
Dear student,
We can proceed as follows:
Moles of HCl = 0.0400 L x 0.800 M=0.0320
Moles of KOH = 5.0 x 10^-3 L x 0.600 M=0.00300
Moles of HCl in excess = 0.0320 - 0.00300=0.0290
Total volume = 0.0450 L
[H+]= 0.0290/ 0.0450=0.644 M
pH = 0.191
Hence moles of KOH = 0.0520 x 0.600 =0.0312
Moles of KOH in excess = 0.0312 - 0.00300=0.0282
Total volume = 0.0920 L
[OH-]= 0.0282/ 0.0920=0.307 M
pOH = 0.514
pH = 14 - 0.514= 13.5
Regards
Sumit
5 years ago
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