How do you calculate [OH^-] in [H_3O^+] = 2.1 times 10^-8 M at 25°C?

Mar 29, 2017

How do we calculate $\left[H {O}^{-}\right]$ IF $\left[{H}_{3} {O}^{+}\right]$ $= 2.1 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$?

Well, [""^(-)OH]=4.78xx10^-7*mol*L^-1.

Explanation:

We know that K_w=[H_3O^+][""^(-)OH]=10^-14, in water at $298 \cdot K$.

We take ${\log}_{10}$ of both SIDES............

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

But ${\log}_{10} {10}^{-} 14 = - 14$ by definition............

And if we DEFINE,
$- {\log}_{10} {K}_{w} = p {K}_{w}$, and $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$

Then................

$p H + p O H = p {K}_{w} = 14$.

You should commit this last result to memory, as it makes the treatment of acidity/basicity straightforward.

So if $\left[{H}_{3} {O}^{+}\right] = 2.1 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$, $p H = - \left(- {\log}_{10} 2.1 \times {10}^{-} 8\right) = - \left(- 7.68\right) = + 7.68$

And thus $p O H = 14 - 7.68 = 6.32$, and $\left[H {O}^{-}\right] = {10}^{- 6.32} = 4.78 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1.$

The solution is slightly BASIC....................