Question

Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

Answer 1

`"pH" = 12.95`

Explanation:

`"KOH"` is a strong base, meaning that it dissociates completely. Therefore, the (balanced) equation is:
`"KOH (aq)" -> "K"^+ "(aq)" + "OH"^-` `(aq)"`

This balanced equation tells us that every mole of `"KOH"` produces one mole of `"OH"^-`.
Therefore, we know that `["KOH"] = ["OH"^-] = 0.09 " M"`

Now, to find the `"pOH"`, we use the formula:
`"pOH" = -log["OH"^-]`

So:
`"pOH" = -log[0.09 " M"] ~~ 1.05`

However, the question wants the `"pH"`, not the `"pOH"`. To find this, we use:
`"pH" = 14 - "pOH"`

So:
`"pH" = 14 - 1.05 = 12.95`

The exact solution to `8` decimal places is `12.95424251`. This is within `0.1%`:

`|(12.95 - 12.95424251)/(12.95424251)| xx 100% ~~ 0.03%`

Hope this helps!