The pH of your solution will be #"13"#.
A solution's pH is all about the concentration of the hydronium ions, or #H_3^(+)O#, present. In your case, the concentration of these ions will be directly influenced by the concentration of the hydroxide ions, or #OH^(-)#, which are added when solid #NaOH# is dissolved in aqueous solution.
The first thing to do is figure out how many moles of #NaOH# you have; this will then lead you to the concentration of the hydroxide ions.
So, if you use sodium hydroxide's molar mass, you can calculate the number of moles by
#"2.00 g NaOH" * "1 mole NaOH"/"40.0 g NaOH" = "0.0500 moles NaOH"#
You can assume that the volume of the solution will be the same as the volume of the water, which means the concentration of sodium hydroxide will be
#C = n/V = "0.0500 moles"/(500 * 10^(-3)"L") = "0.100 M"#
Since sodium hydroxide is a strong base, it will dissociate completely in aqueous solution to give
#NaOH_((aq)) rightleftharpoons Na_((aq))^(+) + OH_((aq))^(-)#
Since 1 mole of #NaOH# produces 1 mole of #OH^(-)#, the concentration of the hydroxide ions will be equal to that of the sodium hydroxide
#C_(OH^(-)) = C_(NaOH) = "0.100 M"#
Finally, to determine pH use
#pH_("solution") = 14 - pOH = 14 - (-log([OH^(-)]))#
#pH_("solution") = 14 - (-log(0.1) ) = 14 - 1 = 13#
Therefore,
#pH_("solution") = 13#
