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Sagot :
We have a solution that pH = 2
We can use this relation which says:
pH + pOH = 14
If we clear pOH from here:
pOH = 14 - pH = 14 - 2 = 12
So, pOH = 12 and pOH could be calculated as:
pOH = - log [OH-] => we clear [OH-] from this,
10^(-pOH) = [OH-] => 10^(-12) = [OH-] => 1x10^-11 mol/L = [OH-]
(mol/L is generally the unit for [OH-])
Answer: [OH-] = 1x10^-11 mol/L
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