A solution contains 0.635 g NaOH in 236 mL solution. What is the equation for the dissociation of NaOH?

I'm guessing that the question is incomplete because you don't really need to know the mass of sodium hydroxide and the volume of the solution in order to write the chemical equation that describes the dissociation of sodium hydroxide.

Sodium hydroxide is a strong base, which implies that it dissociates completely in aqueous solution to produce sodium cations, #"Na"^(+)#, and hydroxide anions, #"OH"^(-)#.

The balanced chemical equation that describes this dissociation looks like this

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

You could mention the fact that the dissociation of sodium hydroxide is water is an exothermic process, which implies that heat is being given off when sodium hydroxide dissolves in water.

This means that you can write

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

Now, you can use the data provided by the problem to determine the molarity of the solution. Molarity is a measure of the number of moles of solute present in #"1 L" = 10^3# #"mL"# of solution.

To convert the mass of sodium hydroxide to grams, use the molar mass of the compound

#0.635 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "0.01588 moles NaOH"#

Next, use the volume of the solution to determine the number of moles of solute present in #10^3# #"mL"# of solution

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01588 moles NaOH"/(236color(red)(cancel(color(black)("mL solution")))) = "0.06729 moles NaOH"#

Since this represents the number of moles of solute present in #10^3# #"mL"# of solution, you can say that its molarity is equal to

#color(darkgreen)(ul(color(black)("molarity = 0.0673 mol L"^(-1))))#

The answer is rounded to three sig figs, the number of sig figs you have for your values.

If you take into account the fact that sodium hydroxide dissociates in #1:1# mole ratios to produce ions in aqueous solution, you can go on to say that the solution has

#["Na"^(+)] = ["OH"^(-)] = "0.0673 mol L"^(-1)#

This will allow you to find the #"pH"# of the solution by using the fact that an aqueous solution at room temperature has

#"pH" = 14 - "pOH"#

which is equivalent to

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

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

In your case, you'd have

#"pH" = 14 + log(0.0673) = 12.83#