# How is the concentration of H+ lons related to the concentration of OH- ions for a substance?

May 15, 2017

Well, we must first specify the water solvent...........and we get under standard conditions, $p H + p O H = 14$.

#### Explanation:

And we know that water undergoes the autoprolysis reaction:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$, we can write,

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$, under standard conditions. Note that ${H}_{3} {O}^{+}$ and ${H}^{+}$ can be used fairly interchangeably. The actual species is possibly a cluster of water molecules with AN extra ${H}^{+}$ associated with it; the proton can tunnel across clusters to give the observed ionic mobility. Likewise, $H {O}^{-}$, is a cluster of water molecules with ONE less ${H}^{+}$; of course we may isolate actual salts of $H {O}^{-}$.

And given ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$, we can take ${\log}_{10}$ of both sides; we remember the definition of the logarithmic function:

log_(10)0.1=log_(10)10^-1=-1; log_(10)1=log_(10)10^0=0;log_(10)10=+1;log_(10)100=2;......................................................... and ${\log}_{10} 1000 = {\log}_{10} {10}^{3} = 3.$

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

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

i.e. $- 14 = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$, multiply each side by $- 1$, and thus we get......

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

And thus our defining relationship, the goal of all the previous rigmarole:

$p H + p O H = 14$

You will not have to derive this function, but you will use it. Equivalently, in aqueous solution, $p {K}_{a} + p {K}_{b} = 14$.