What will be the OH^{-} concentration in an a | vedclass

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(B) Given that the $pH$ of the solution is $8$.We know that $pH + pOH = 14$ at $25^{\circ}C$.Therefore,$pOH = 14 - pH = 14 - 8 = 6$.The relationship b

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$1 	imes 10^{-6} \, M$

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(B) Given that the $pH$ of the solution is $8$.We know that $pH + pOH = 14$ at $25^{\circ}C$.Therefore,$pOH = 14 - pH = 14 - 8 = 6$.The relationship between $pOH$ and $[OH^{-}]$ is given by $pOH = -\log[OH^{-}]$.So,$6 = -\log[OH^{-}]$,which implies $\log[OH^{-}] = -6$.Taking the antilog,we get $[OH^{-}] = 10^{-6} \, M$ or $1 	imes 10^{-6} \, M$.

What will be the $OH^{-}$ concentration in an aqueous solution having $pH = 8$?

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