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avo 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..
2 years ago

Answers : (1)

Sumit Majumdar
IIT Delhi
askIITians Faculty
132 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
2 years ago
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