Question #ca771

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Question #ca771

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Answer 1 · 2017-05-24T03:00:57

pH = 11.87

Explanation:

We know if we have a given $[O H^{-}]$ $[OH^(-)]$ , then we can find $pOH$ $"pOH"$ which is just the $-log of the [O H^{-}] .$ $"-log of the" [OH^(-)].$

Find $pOH$ $"pOH"$

$- log [O H^{-}] \to - log [7.5 \cdot 10^{- 3} M] \to 2.13$ $-log[OH^(-)]->-log[7.5 * 10^-3 M]->2.13$

$a a a$ $color(white)(aaa)$

We also know from the relationship that

$a a a a a a pH+pOH = 14 at 25^{\circ} Celsius$ $color(white)(aaaaaa)color(green)("pH+pOH = 14 at "25^@"Celsius"$

so from the $pOH$ $"pOH"$ we can find the $pH$ $"pH"$ .

$a a a$ $color(white)(aaa)$

Find $pH$ $"pH"$ from $pOH$ $"pOH"$

$pH+pOH = 14$ $"pH+pOH = 14"$
$pH =14-2.13$ $"pH =14-2.13"$
$pH = 11.87$ $"pH = 11.87"$

Answer: pH = 11.87

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Question #ca771

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