# What is the pH of pure water at 25.0 °C if the Kw at this temperature is 1.0 x 10-^14?

##### 1 Answer
Jun 9, 2018

$p H = 7.00$.

#### Explanation:

Note: It's generally pretty well known that the $p H$ of pure water at $\text{25 ºC}$ is $7$, but here's what we'd do to get to find that $p H$ if we didn't know this:

To find $p H$, we'll need to use two formulas:

1. ${K}_{w} = \left[{H}^{+}\right] \times \left[O {H}^{-}\right]$
2. $p H = - \log \left[{H}^{+}\right]$

First, we need to find the concentration of ${H}^{+}$ ions in water at this temperature.

We know that pure water is neutral, and that in neutral solutions, the concentration of ${H}^{+}$ equals the concentration of $O {H}^{-}$.

So, effectively:

${K}_{w} = \left[{H}^{+}\right] \times \left[O {H}^{-}\right]$
${K}_{w} = \left[{H}^{+}\right] \times \left[{H}^{+}\right]$ because $\left[{H}^{+}\right] = \left[O {H}^{-}\right]$ in neutral solution

To find the concentration of ${H}^{+}$ ions now, we'll just need to square root ${K}_{w}$. The question tells us that ${K}_{w} = 1.0 \times {10}^{- 14}$, so:

$1.0 \times {10}^{- 14} = \left[{H}^{+}\right] \times \left[{H}^{+}\right]$

$\left[{H}^{+}\right] = \sqrt{1.0 \times {10}^{- 14}} = 1.0 \times {10}^{- 7} M$

Now, we can just use our second equation to calculate the $p H$:

$p H = - \log \left[{H}^{+}\right]$
$p H = - \log \left(1.0 \times {10}^{- 7} M\right) = 7.00$

(The number of decimal places in our $p H$ value needs to be the same as the number of significant figures in our $\left[{H}^{+}\right]$ value.)