# Question 4a1f3

Apr 4, 2017

$p O H = 4.255 \cdot {10}^{- 13}$

#### Explanation:

From the concentration of HCl, we can calculate the pH using the following rule:
$p H = - \log \left[{H}_{\text{3"O^"+}}\right]$
We obtain:
$p H = - \log \left[0.0235\right] = 1.6289$

Now we use the formula
$p H + p O H = 14$
We can write that like this:
$p O H = 14 - p H$

Since pH=1.6289, we calculate pOH
$p O H = 14 - 1.6289 = 12.37$

Now the $\left[O {H}^{\text{-}}\right]$ can be calculated from the pOH by using this formula: $\left[O {H}^{\text{-}}\right] = {10}^{- p O H}$
We obtain: $\left[O {H}^{\text{-}}\right] = {10}^{- 12.37} = 4.255 \cdot {10}^{- 13}$

----------------------------------------------------------

How is $p H + p O H = 14$ established?
In water, the following (ionization) reaction occurs:
$2 {H}_{\text{2"O -> H_"3"O^"+" + OH^"-}}$

Therefore the equilibrium can be written like
K_"c"=([H_"3"O^"+"]*[OH^"-"])/[H_"2"O]
Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:
K_"c"=[H_"3"O^"+"]*[OH^"-"]

The ${K}_{\text{c}}$ in this equation represents a special number because we talk about the ionisation of water. Therefore we denote ${K}_{\text{c}}$ as ${K}_{\text{w}}$. The value of the ${K}_{\text{w}}$ is measured at 25°C.
K_"w" (25°C) = 1*10^(-14)
This means we can say:
K_"c"=K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)#

To get from the $\left[{H}_{\text{3"O^"+}}\right]$ (concentration ${H}_{\text{3"O^"+}}$) to the pH, we use the following formula:
$p H = - \log \left[{H}_{\text{3"O^"+}}\right]$
The same is true for the $\left[O {H}^{\text{-}}\right]$, since we define pOH as
$p O H = - \log \left[O {H}^{\text{-}}\right]$

Now if we take the Log from both sides of the ${K}_{\text{w}}$ equation, we get:
$\log \left(1 \cdot {10}^{- 14}\right) = \log \left(\left[{H}_{\text{3"O}}\right] \cdot \left[O {H}^{-}\right]\right)$
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
$\log \left({10}^{- 14}\right) = \log \left[{H}_{\text{3"O}}\right] + \log \left[O {H}^{-}\right]$

And now we can use the definitions of pOH and OH! We get:
$\log \left({10}^{- 14}\right) = - p H - p O H$
with $\log \left({10}^{- 14}\right) = - 14$ we get the function
$- p H - p O H = - 14$
Which is the same as
$p H + p O H = 14$