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
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: