The pH of a solution is 5.6. What are the hydrogen and hydroxide concentrations?

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Answer 1 · 2016-05-13T21:33:52

$p H = - log [H_{3} O^{+}]$ $pH =-log[H_3O^+]$

$[H_{3} O^{+}] \cdot [O H^{-}] = 10^{- 14}$ $[H_3O^+]*[OH^-]=10^-14$

The pH of a given solution is determined using the following formula:

$p H = - log [H_{3} O^{+}]$ $pH =-log[H_3O^+]$

To find the $[H_{3} O^{+}]$ $[H_3O^+]$ the formula must be manipulated as follows:

$log [H_{3} O^{+}] = - p H$ $log[H_3O^+]= -pH$

$[H_{3} O^{+}] = 10^{- p H}$ $[H_3O^+]=10^(-pH)$

Plugging the value of the pH in the equation gives:

$[H_{3} O^{+}] = 10^{- 5.6}$ $[H_3O^+]=10^-5.6$

$[H_{3} O^{+}] = 10^{0.4} \cdot 10^{- 6}$ $[H_3O^+]=10^0.4*10^(-6)$

$[H_{3} O^{+}] = 2.51 \cdot 10^{- 6} M$ $[H_3O^+]=2.51*10^-6 M$

In an aqueous solution,

$[H_{3} O^{+}] \cdot [O H^{-}] = 10^{- 14}$ $[H_3O^+]*[OH^-]=10^-14$

$[O H^{-}] = \frac{10^{- 14}}{[H_{3} O^{+}]}$ $[OH^-]=(10^-14)/([H_3O^+])$

$[O H^{-}] = \frac{10^{- 14}}{2.51 \cdot 10^{- 6}} = 3.98 \cdot 10^{- 9} M$ $[OH^-]=(10^-14)/(2.51*10^-6) =3.98*10^-9 M$