Hi friends ,
Please solve a problem
PH of 0.05 M Ba(OH)2 ?
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Hi friend,
Its ph will be 13. SOLUTION: Ba(OH)2 - Ba2+ 2OH-
0.05 M X 2= 0.10M pOH= -log(0.10 M) = 1.00
pH = 14-pOH= 14 - 1.00=13
HOPE THIS HELPS YOU:-))
Its ph will be 13. SOLUTION: Ba(OH)2 - Ba2+ 2OH-
0.05 M X 2= 0.10M pOH= -log(0.10 M) = 1.00
pH = 14-pOH= 14 - 1.00=13
HOPE THIS HELPS YOU:-))
Rahul00000:
Thank you
:-)
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