Question

What is the [`OH^-`] in a solution of pH 3.00?

Answer 1

`[OH^"-"]=10^(-11)`

Explanation:

For this answer, we use the formula:
`pH+pOH =14`
We can write that like this:
`pOH =14-pH`

Since pH=3 is given, we calculate pOH
`pOH=14-3=11`

Now the `[OH^"-"]` can be calculated from the pOH by using this formula: `[OH^"-"]=10^(-pOH)`
We obtain: `[OH^"-"]=10^(-11)`
For more information about pH, check here!

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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 our function
`-pH-pOH =-14`
Which is the same as
`pH+pOH=14`