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

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