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

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.