Question #881f7

The first thing to do here is to figure out the molarity of the sodium hydroxide solution.

To do that, use the molar mass of sodium hydroxide to calculate how man moles of solute you have in your sample

#100 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "2.50 moles NaOH"#

Now, the molarity of the solution tells you the number of moles of solute present per liter of solution.

In your case, you know that #"500 mL"# of solution contain #2.50# moles of sodium hydroxide, the solute. You can thus say that #"1 L"# of this solution will contain

#1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "2.50 moles NaOH"/(500color(red)(cancel(color(black)("mL solution")))) = "5.0 moles NaOH"#

Therefore, the molarity of the solution will be equal to

#"molarity = 5.0 mol L"^(-1)#

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

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

As you can see, every #1# mole of sodium hydroxide that dissolves in water produces #1# mole of hydroxide anions. In other words, you have

#["OH"^(-)] = ["NaOH"] = "5.0 mol L"^(-1)#

Now, you know that you have

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

Moreover, an aqueous solution at room temperature has

#"pH + pOH = 14"#

You can thus say that the #"pH"# of the solution will be equal to

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

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

In your case, you will have

#"pH" = 14 + log(5.0) = color(darkgreen)(ul(color(black)(14.7)))#

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

Notice that you have

#"pH" > 14#

This is not really an issue because any concentration of hydroxide anions that is #"> 1.0 M"# will result in a #"pH"# value that is #> 14#.