If $[H_{3} O^{+}] = 4.5 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=4.5xx10^-3molL^-1$ , what are $p H$ $pH$ , and $p O H$ $pOH$ , and $[H O^{-}]$ $[HO^-]$ ?

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Answer 1 · 2017-05-19T04:43:20

$pH=-log_(10)[H_3O^+]...............$

$(i)$ $pH=-log_10(4.5xx10^-3)=-(-2.35)=2.35$ .

For $pOH$ , we recall that $14=pOH+pH$ for protonolysis in water...

$(ii)$ $pOH=-log_10(3.67xx10^-5)=-(-2.35)=4.44$ ; and thus, from the expression above, $pH=9.56$ .

I will let you do the last problem.

If [H_3O^+]=4.5xx10^-3*mol*L^-1[H3O+]=4.5×10−3⋅mol⋅L−1[H_3O^+]=4.5xx10^-3*mol*L^-1, what are pHpHpH, and pOHpOHpOH, and [HO^-][HO−][HO^-]?