What is pH if (i) [H_3O^+]=5.6xx10^-5*mol*L^-1; and (ii) if [HO^-]=10^-7*mol*L^-1?

Jun 25, 2017

$\left(i\right)$ $p H = 4.25 \ldots \ldots \ldots \ldots \ldots . .$

Explanation:

By definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and thus for (i)....

$p H = - {\log}_{10} \left(5.6 \times {10}^{-} 5\right) = - \left(- 4.25\right) = 4.25$ (if I have counted up the zeroes properly........)

But since $\left[{H}_{3} {O}^{+}\right] \times \left[H {O}^{-}\right] = {10}^{-} 14$, we can take ${\log}_{10}$ of both sides, and gets........

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = {\log}_{10} {10}^{-} 14 = - 14$

Equivalently, multiplying each side by $- 1$............

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\textcolor{red}{p H}} {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{\textcolor{b l u e}{p O H}} = + 14$

And thus our defining relationship......

$\textcolor{red}{p H} + \textcolor{b l u e}{p O H} = 14$

And so (ii) if $\left[H {O}^{-}\right] = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$, $p O H = 7$ (why?), and $p H = 7$. Agreed?