Calulate the {OH-} if the pH is 11.6?



Answer:
pH = 11.6
=>pOH = 14-11.6 = 2.4
=> - log[OH(-)] = 2.4

=> [OH(-)] =10^(-2.4) gm ion/L
pH=11.6
pOH=2.4
[OH]=0.00398=3.98x10^-3
POH is 2.4

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