# What is the concentration of hydroxide ions in pure water at 30.0°C, if K_w at this temperature is 1.47*10^-14?

Feb 12, 2017

We interrogate the equilibrium: "2H_2O(l) rightleftharpoons H_3O^(+) + HO^-.

$\left[H {O}^{-}\right] = 1.21 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We interrogate the equilibrium:

"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-,

where the ion product, ${K}_{w} = 1.47 \times {10}^{-} 14$, at $303 \cdot K$. Note that this is slightly higher than ${K}_{w} = {10}^{-} 14$ at $298 \cdot K$, and given that this is a bond breaking reaction this is perhaps reasonable.

From the equation, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = 1.47 \times {10}^{-} 14$.

Given that this is a neutral solution, $\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right]$.

Thus ${\left[H {O}^{-}\right]}^{2} = 1.47 \times {10}^{-} 14$

And $\left[H {O}^{-}\right] = \sqrt{1.47 \times {10}^{-} 14} = 1.21 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$