We're asked to find the molarity of the "NaOH" solution given some titration values.
Let's first write the balanced chemical equation for this reaction:
2"NaOH"(aq) + "H"_2"SO"_4(aq) rarr "Na"_2"SO"_4 (aq) + 2"H"_2"O" (l)
We can calculate the number of moles of "H"_2"SO"_4 used by using the molarity equation:
"mol solute" = ("molarity")("L soln")
We'll convert the volume to liters to use this equation:
"mol H"_2"SO"_4 = (0.5"mol"/(cancel("L")))(0.0205cancel("L")) = color(red)(0.0103 color(red)("mol H"_2"SO"_4
Using the coefficients of the chemical equation, we can determine thee relative number of moles of "NaOH" used:
color(red)(0.0103)cancel(color(red)("mol H"_2"SO"_4))((2color(white)(l)"mol NaOH")/(1cancel("mol H"_2"SO"_4))) = 0.0205 "mol NaOH"
Lastly, we'll use the given volume of the "NaOH" solution and the mole value to calculate the molarity of the solution:
"molarity" = "mol solute"/"L soln"
= (0.0205color(white)(l)"mol NaOH")/(0.020color(white)(l)"L") = color(blue)(1M
which I'll leave to one significant figure I suppose (real value = 1.025M).
