# A solution has a pOH of 6.39. What is the solutions pH? (You must answer to the correct number of s.d.'s)

May 28, 2017

$p H = 14 - 6.39 = 7.61$

#### Explanation:

For the acid-base equilibrium that operates in water.....

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

We know that at under standard conditions, $298 \cdot K , \text{1 atmosphere}$.............

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

This defines the acid-base equilibrium, the autoprotolysis of water.

We can take ${\log}_{10}$ of BOTH sides........

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

And on rearrangement,

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

Given that ${K}_{w} = {10}^{-} 14$, and $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, then BY DEFINITION, ${\underbrace{- {\log}_{10} {K}_{w}}}_{p {K}_{w}} = - {\log}_{10} {10}^{- 14} = - \left(- 14\right) = 14$, and the defining relationship, which you may not have to derive, but WILL have to remember,

$14 = p H + p O H$

And given that $p O H = 6.39$, this means that $\left[H {O}^{-}\right] = {10}^{- 6.39} \cdot m o l \cdot {L}^{-} 1$, and $p H = 14 - 6.39 = 7.61$, and (FINALLY) $\left[{H}_{3} {O}^{+}\right] = {10}^{- 7.61} \cdot m o l \cdot {L}^{-} 1 = 2.4 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1.$

Now this might seem a lot of work, but only because I derived the equation. You must be able to use the relationship...........

$14 = p H + p O H$

These logarithmic terms were introduced back in the day, before the advent of electronic calculators. Log tables, printed values of ${\log}_{10}$ and ${\log}_{e}$ were widely used by scientists, engineers, and by students of course.

May 28, 2017

pOH 6.39 gives pH =7.61

#### Explanation:

For this question, you can use the rule:
$p H + p O H = 14$

Rewritten this gives us:
$p H = 14 - p O H$
$p H = 14 - 6.39$
$p H = 7.61$

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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
$p H + p O H = 14$