Question

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

Answer 1

`pH=14-6.39=7.61`

Explanation:

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

`2H_2O rightleftharpoons H_3O^+ + HO^-`

We know that at under standard conditions, `298*K, "1 atmosphere"`.............

`K_w=[H_3O^+][HO^-]=10^-14`

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

We can take `log_10` of BOTH sides........

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]`

And on rearrangement,

`underbrace(-log_10[H_3O^+])_(pH)-underbrace(log_10[HO^-])_(pOH)=underbrace(-log_10K_w)_(pK_w)`

Given that `K_w=10^-14`, and `pH=-log_10[H_3O^+]`, then BY DEFINITION, `underbrace(-log_10K_w)_(pK_w)=-log_(10)10^(-14)=-(-14)=14`, and the defining relationship, which you may not have to derive, but WILL have to remember,

`14=pH +pOH`

And given that `pOH=6.39`, this means that `[HO^-]=10^(-6.39)*mol*L^-1`, and `pH=14-6.39=7.61`, and (FINALLY) `[H_3O^+]=10^(-7.61)*mol*L^-1=2.4xx10^-8*mol*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=pH +pOH`

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.

Answer 2

pOH 6.39 gives pH =7.61

Explanation:

For this question, you can use the rule:
`pH +pOH=14`

Rewritten this gives us:
`pH=14-pOH`
`pH=14-6.39`
`pH=7.61`

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How is `pH+pOH =14` established?

In water, the following (ionization) reaction occurs:
`2 H_"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_"c"` in this equation represents a special number because we talk about the ionisation of water. Therefore we denote `K_"c"` as `K_"w"`. The value of the `K_"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 `[H_"3"O^"+"]` (concentration `H_"3"O^"+"`) to the pH, we use the following formula:
`pH= - log[H_"3"O^"+"]`
The same is true for the `[OH^"-"]`, since we define pOH as
`pOH=- log[OH^"-"]`

Now if we take the Log from both sides of the `K_"w"` equation, we get:
`log(1*10^(-14))=log([H_"3"O"]*[OH^-])`
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
`log(10^(-14))=log[H_"3"O"]+log[OH^-]`

And now we can use the definitions of pOH and OH! We get:
`log(10^(-14))=-pH - pOH`
with `log(10^(-14))=-14` we get the function
`-pH-pOH =-14`
Which is the same as
`pH+pOH=14`