The pH of a 2M solution of NaOH is...? (please include the complete solution of the problem)

For starters, you should know that an aqueous solution at room temperature has

#color(blue)(ul(color(black)("pH + pOH = 14")))#

This means that you express the #"pH"# of the solution in terms of its #"pOH"# by writing

#"pH" = 14 - "pOH"#

Now, the #"pOH"# of the solution is defined as the negative log base #10# of the concentration of hydroxide anions

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

This means that the #"pH"# of the solution can be expressed as

#"pH" = 14 - [- log(["OH"^(-)])]#

#"pH" = 14 + log(["OH"^(-)])" "color(darkorange)("(*)")#

As you know, sodium hydroxide is a strong base, which implies that it ionizes completely in aqueous solution to produce sodium cations and hydroxide anions.

You can thus say that

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

Since every #1# mole of sodium hydroxide that is dissolved in water produces #1# mole of hydroxide anions, your solution will have

#["OH"^(-)] = ["NaOH"]#

In your case, this is equal to

#["OH"^(-)] = "2 M"#

Plug this into equation #color(orange)("(*)")# and calculate the #"pH"# of the solution

#"pH" = 14 + log(2) = color(darkgreen)(ul(color(black)(14.3)))#

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the solution.

As a final note, don't get confused by the fact that you have

#"pH" > 14#

This is what you'll always get for basic solutions that have

#["OH"^(-)] > 1#

and it means that #"pOH" < 0#.