# What is [HO^-] of a solution that is 6.80xx10^-2*mol*L^-1 with respect to HBr(aq)?

Mar 13, 2017

$\left[H {O}^{-}\right] = 1.47 \times {10}^{-} 13 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We interrogate the hydrolysis reaction of water under standard conditions:

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

The extent of this equilibrium is known very precisely, and we find (under standard conditions) that:

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

The given equation is an $\text{equation}$ (duhh!); we may divide it, multiply it, add to it, PROVIDED that we do it consistently to both left hand and right sides of the equation. One thing we can do is to take $\text{logarithms to the base 10}$ of both sides, and follow the standard rules of arithmetic:

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

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

Upon rearrangement:

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

Now using the definition of a logarithm, when I write ${\log}_{a} b = c$, this means that ${a}^{c} = b$. Perhaps you should review the logarithmic function in your mathematics text.

So the left hand side of the equation, $= - {\log}_{10} \left({10}^{-} 14\right) = - 14$, i.e.

$- {\log}_{10} \left({10}^{-} 14\right) = 14$

So I can rearrange the given equation to yield:

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

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

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and

$- {\log}_{10} \left[H {O}^{-}\right] = p O H$.

And so finally, we can write under standard conditions that $p H + p O H = 14$.

And with all this done, we can (FINALLY!) address the given question (which I have almost forgotten).

We have $\left[{H}_{3} {O}^{+}\right] = 6.80 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$, why? Well because the $H B r$ acid dissociates completely in aqueous solution according to the following equation:

$H B r \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + B {r}^{-}$

And so $p H = - {\log}_{10} \left(6.80 \times {10}^{-} 2\right) = 1.17$,

and $p O H = 12.83$.

And so $\left[H {O}^{-}\right] = {10}^{- 12.83} = 1.471 \times {10}^{-} 13 \cdot m o l \cdot {L}^{-} 1$.

I acknowledge that this was a lot of work for a small problem. All we have done here is to use the logarithmic function appropriately.

The key equation is $p H + p O H = 14$.