Question

OH- Concentration of 0.015 M HCl?

Answer 1

`approx 6.61 times 10^-13 mol dm^-3`

Explanation:

Assuming that this is done under standard conditions we can use the following relationship:

`pH+pOH=14` (at 25 degrees Celcius)

This is crucial since it connects `pH` and `pOH` of a given solution- lets say that the`pH` is 1, then the `pOH` will be 13.

Also remember that:

`pH=-log[H_3O^+]`

`pOH=-log[OH^-]`

In this problem, We can assume that `HCl` is completely ionized in water because `HCl` is a strong acid so the `H_3O^+` concentration will also be `0.015 mol dm^-3`

Then using the equation to find the pH of the given solution:

`pH=-log[H_3O^+]`

`pH=-log[0.015]`

`pH approx 1.82`

`1.82 + pOH=14`

`pOH approx 12.18`

Now we must reverse-engineer the process.

If:

`pOH=-log[OH^-]`

Then:

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

Hence:

`10^-(12.18)=[OH^-]`

`[OH^-]=10^-(12.18)approx 6.61 times 10^-13 mol dm^-3`