Please calculate the ph at 25 degrees C of 0.080 mol `dm^(-3)` KOH Please provide the steps. Thanks!

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PH measures the relative acidity (or basicity) of a certain solution which is due to the compounds that are present in the solution. Approximately at 25 degrees Celsius, the formula in getting the pH is:

`pH = -log [H^(+)]`

Where `[H^(+)] ` = concentration of the acid

For basic solutions, we use:

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

Where `[OH^(-)]` = concentration of the basic solution

The relationship of the two equations can be expressed as:

`pH + pOH = 14`

We know that KOH is a base, so we should be using the latter formula.

`[OH^(-)] = concentration of KOH`

`[OH^(-)] = 0.080 (mol)/(dm^(3))`

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

`pOH = -log [0.080]`

`pOH =1.097`

`pH + pOH = 14`

`pH = 14 - pOH`

`pH = 14 -1.097`

`pH = 12.903` -> final answer 

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