# If [HO^-]=6.0xx10^-10*mol*L^-1, what is pH for this solution?

Oct 31, 2016

$\left[{H}^{+}\right] \left(\text{or } \left[{H}_{3} {O}^{+}\right]\right) = {10}^{- 4.78} \cdot m o l \cdot {L}^{-} 1$, $p H = 4.78$, $p O H = 9.22$

#### Explanation:

In aqueous solution at $298 \cdot K$, water undergoes the following equilibrium reaction:

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

Alternatively, and perhaps now more commonly we would write:

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

This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:

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

If we take negative logarithms to the base 10 of both sides we get:

$p {K}_{w} = 14 = p O H + p H$, where the $p H$ functions means $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, etc.

We are given $\left[H {O}^{-}\right] = 6.0 \times {10}^{-} 10$, so $p O H = - {\log}_{10} \left(6.0 \times {10}^{-} 10\right) = - \left(- 9.22\right) = 9.22$, and thus $p H = 14 - 9.22 = 4.78$.

Is this solution acidic or basic?

The given reaction occurs at $298 \cdot K$. Given that this is a bond-breaking reaction, how do you think $p {K}_{w}$ would evolve at $323 \cdot K$ or $373 \cdot K$?