......answer????? .......
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PH = - log[ H+]
= - log( 10^-6)
= 6
so
PoH = 14 -6
= 8
now
POH = - log[ oH-]
8 = - log[OH-]
[OH-] = 10^-8
##hope it can help you #$
= - log( 10^-6)
= 6
so
PoH = 14 -6
= 8
now
POH = - log[ oH-]
8 = - log[OH-]
[OH-] = 10^-8
##hope it can help you #$
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