Answer :
Answer: option C) 5x10−6 M
Explanation:
Recall that pH + pOH = 14
9.0 + pOH = 14
pOH = 14 - 9.0
pOH = 5.0
Then, find the concentration of aqueous Ba(OH)2 that yields a pOH of 5.0
Ba(OH)2 = Ba2+ + 2OH-
Since pOH = - log(OH-)
5.0 = - 2 x log(OH-)
2 x (OH-) = Antilog (-5.0)
2 x (OH-) = 0.00001
(OH-) = 0.00001/2
(OH-) = 0.000005
Then, 0.000005 in standard form = 5x10−6 M
[tex]5\times10^{-6} M[/tex] is the concentration of aqueous barium hydroxide.
Relationship between pH and pOH
pH + pOH = 14
We can calculate the pOH from the above equation,
9.0 + pOH = 14
pOH = 14 - 9.0
pOH = 5.0
Then, find the concentration of aqueous [tex]Ba(OH)_{2}[/tex] that yields a pOH of 5.0
[tex]Ba(OH)_{2} = Ba^{2+} + 2OH^{-}[/tex]
Calculate the concentration of hydroxide ion as follows,
[tex]pOH=-log(OH^{-} )[/tex]
[tex]5.0 = - 2\times log(OH^{-} )\\2 \times (OH^{-} ) = Antilog (-5.0)\\2\times(OH^{-} ) = 0.00001\\(OH^{-} ) = 0.00001/2\\(OH^{-} ) = 0.000005\\=5\times10^{-6} M[/tex]
So, the concentration of aqueous barium hydroxide is [tex]5\times10^{-6} M[/tex].
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