Question

The [OH-] of an aqueous solution is `4.62 * 10^-4` M. What is the pH of this solution?

Answer 1

`"pH" = 10.66`

Explanation:

For pure water at `25^@"C"`, the concentration of hydronium ions, `"H"_3"O"^(+)`, is equal to the concentration of hydroxide ions, `"OH"^(-)`.

More specifically, water undergoes a self-ionization reaction that results in the formation of equal concentrations of hydronium and hydroxide anions.

`2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)`

At room temperature, the self-ionization constant of water is equal to

`K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)`

This means that neutral water at this temperature will have

`["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"`

As you know, pH and pOH are defined as

`"pH" = - log( ["H"_3"O"^(+)])`

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

and have the following relationship

`"pH" + "pOH" = 14`

In your case, the concentration of hydroxide ions is bigger than `10^(-7)"M"`, which tells you that you're dealing with a basic solution and that you can expect the pH of the water to be higher than `7`.

A pH equal to `7` is characteristic of a neutral aqueous solution at room temperature.

So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first

`"pOH" = - log( 4.62 * 10^(-4)) = 3.34`

This means that the solution's pH will be

`"pH" = 14 - "pOH"`

`"pH" = 14 - 3.34 = color(green)(10.66)`

Indeed, the pH is higher than `7`, which confirms that you're dealing with a basic solution.