Question

Seawater has a pH of 8.100. What is the concentration of OH–?

Answer 1

`10^-5.9 approx 1.26 times 10^-6 mol dm^-3`

Explanation:

If the `pH` is 8.1, and we assume that this is measured under standard conditions, we can use the relationship:

`pH+pOH=pK_w`

At 25 degrees celcius, `pK_w=14`
Where `K_w` is the dissociation constant for water- `1.0 times 10^-14`,(At 25 degrees C) but `pK_w` is the negative logarithm of `K_w`.

`pK_w=-log_10[K_w]`

From this, we can convert `pH`, the measure of `H_3O^+` ions, into `pOH`, the measure of `OH^-` ions in the seawater:

`pH+pOH=14`
`8.1+ pOH=14`
`pOH=5.9`

Then we know that:
`pOH=-log_10[OH^-]`
So to rearrange the equation to solve for `[OH^-]`:

`10^(-pOH)= [OH^-]`
Hence:

`10^(-5.9)=[OH^-] approx 1.26 times 10^-6 mol dm^-3`

Answer 2

`1.26*10^-6 \ "M"`

Explanation:

Well, the `"pH"` of the seawater is `8.1`, and so its `"pOH"` is:

`"pOH"=14-8.1`

`=5.9`

The `"pOH"` of a substance is related through the equation,

`"pOH"=-log[OH^-]`

• `[OH^-]` is the hydroxide ion concentration in terms of molarity.

And so, we got:

`-log[OH^-]=5.9`

`[OH^-]=10^-5.9`

`~~1.26*10^-6 \ "M"`